Samacheer Kalvi Class 11 Maths Solutions Chapter 9 Limits and Continuity Exercise 9.5

 

Question 1. Prove that \( f(x) = 2x^2 + 3x - 5 \) is continuous at all points in R.
Answer:Let the function be \( f(x) = 2x^2 + 3x - 5 \). This function is a polynomial, which means it is defined for every real number \( x \). To check continuity at any point, let's pick an arbitrary real number, say \( x_0 \). First, find the function's value at \( x_0 \):
\( f(x_0) = 2x_0^2 + 3x_0 - 5 \) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} (2x^2 + 3x - 5) \)
Since polynomials are continuous, we can directly substitute \( x_0 \):
\( \lim_{x \to x_0} f(x) = 2x_0^2 + 3x_0 - 5 \) We can see that the limit of \( f(x) \) as \( x \) approaches \( x_0 \) is equal to \( f(x_0) \). This holds true for any real number \( x_0 \). Therefore, the function \( f(x) \) is continuous at all points in the set of real numbers, R.
In simple words: This function is always smooth and has no breaks or jumps. For any point you pick, the function's value at that point is exactly what the graph approaches from both sides. This is because it's a simple polynomial.

🎯 Exam Tip: Remember that all polynomial functions are continuous everywhere in their domain, which is usually all real numbers. State this property as a reason for continuity.

 

Question 2. Examine the continuity of the following
(i) \( x + \sin x \)
Answer:Let \( f(x) = x + \sin x \). The function \( f(x) \) is defined for all real numbers \( x \). This is because both \( x \) and \( \sin x \) are individually defined for all real numbers. Let \( x_0 \) be an arbitrary real number. First, find the value of the function at \( x_0 \):
\( f(x_0) = x_0 + \sin x_0 \) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} (x + \sin x) \)
Since the limit of a sum is the sum of the limits, and both \( x \) and \( \sin x \) are continuous functions:
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} x + \lim_{x \to x_0} \sin x = x_0 + \sin x_0 \) Because \( \lim_{x \to x_0} f(x) = f(x_0) \), the function \( f(x) \) is continuous at all points in R.
In simple words: The function \( x + \sin x \) is continuous everywhere. Both \( x \) (a straight line) and \( \sin x \) (a smooth wave) are continuous on their own, so their sum is also continuous.

🎯 Exam Tip: The sum, difference, product, and quotient of continuous functions are also continuous within their common domain. Use this property to quickly establish continuity for combined functions.

 

Question 2. Examine the continuity of the following
(ii) \( x^2 \cos x \)
Answer:Let \( f(x) = x^2 \cos x \). The function \( f(x) \) is defined for all real numbers \( x \). This is because both \( x^2 \) and \( \cos x \) are individually defined for all real numbers. Let \( x_0 \) be an arbitrary real number. First, find the value of the function at \( x_0 \):
\( f(x_0) = x_0^2 \cos x_0 \) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} (x^2 \cos x) \)
Since the limit of a product is the product of the limits, and both \( x^2 \) and \( \cos x \) are continuous functions:
\( \lim_{x \to x_0} f(x) = (\lim_{x \to x_0} x^2) (\lim_{x \to x_0} \cos x) = x_0^2 \cos x_0 \) Because \( \lim_{x \to x_0} f(x) = f(x_0) \), the function \( f(x) \) is continuous at all points in R.
In simple words: The function \( x^2 \cos x \) is continuous everywhere. Since \( x^2 \) (a parabola) and \( \cos x \) (a smooth wave) are continuous by themselves, their product will also be continuous.

🎯 Exam Tip: Remember that polynomial functions like \( x^2 \) and trigonometric functions like \( \cos x \) are continuous on R. Their product will also be continuous on R.

 

Question 2. Examine the continuity of the following
(iii) \( e^x \tan x \)
Answer:Let \( f(x) = e^x \tan x \). The function \( e^x \) is continuous for all real numbers. However, the function \( \tan x \) is not defined at points where \( \cos x = 0 \), which occurs at \( x = (2n + 1)\frac{\pi}{2} \) for any integer \( n \). Therefore, \( f(x) \) is defined and continuous on the domain \( R - \{ (2n + 1)\frac{\pi}{2} : n \in Z \} \). Let \( x_0 \) be an arbitrary point in this domain. First, find the value of the function at \( x_0 \):
\( f(x_0) = e^{x_0} \tan x_0 \) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} (e^x \tan x) \)
Since both \( e^x \) and \( \tan x \) are continuous in the chosen domain, we can write:
\( \lim_{x \to x_0} f(x) = (\lim_{x \to x_0} e^x) (\lim_{x \to x_0} \tan x) = e^{x_0} \tan x_0 \) Because \( \lim_{x \to x_0} f(x) = f(x_0) \), the function \( f(x) \) is continuous at all points within its domain.
In simple words: The function \( e^x \tan x \) is continuous everywhere except where \( \tan x \) has breaks, which are at \( \frac{\pi}{2}, \frac{3\pi}{2}, -\frac{\pi}{2} \), and so on. These are the points where the tangent function goes to infinity.

🎯 Exam Tip: When dealing with products or quotients of functions, always remember to identify the domain of each component function and the overall function. Discontinuities often arise where a function is undefined.

 

Question 2. Examine the continuity of the following
(iv) \( e^{2x} + x^2 \)
Answer:Let \( f(x) = e^{2x} + x^2 \). The function \( e^{2x} \) is an exponential function, which is defined and continuous for all real numbers. The function \( x^2 \) is a polynomial, also defined and continuous for all real numbers. Since \( f(x) \) is the sum of two functions that are continuous everywhere, \( f(x) \) is also defined and continuous for all real numbers \( x \). Let \( x_0 \) be an arbitrary real number. First, find the value of the function at \( x_0 \):
\( f(x_0) = e^{2x_0} + x_0^2 \) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} (e^{2x} + x^2) \)
Since both \( e^{2x} \) and \( x^2 \) are continuous, we can substitute directly:
\( \lim_{x \to x_0} f(x) = e^{2x_0} + x_0^2 \) Because \( \lim_{x \to x_0} f(x) = f(x_0) \), the function \( f(x) \) is continuous at all points in R.
In simple words: This function is continuous everywhere. Both \( e^{2x} \) (an exponential curve) and \( x^2 \) (a parabola) are always smooth without any breaks, so adding them together results in another smooth, continuous function.

🎯 Exam Tip: Exponential functions \( e^{ax} \) and polynomial functions are fundamental examples of functions that are continuous across all real numbers. Their sum will also maintain this continuity.

 

Question 2. Examine the continuity of the following
(v) \( x \log x \)
Answer:Let \( f(x) = x \log x \). The function \( x \) is defined and continuous for all real numbers. However, the natural logarithm function \( \log x \) is only defined for \( x > 0 \). Therefore, the function \( f(x) \) is defined and continuous only on the open interval \( (0, \infty) \). Let \( x_0 \) be an arbitrary point in \( (0, \infty) \). First, find the value of the function at \( x_0 \):
\( f(x_0) = x_0 \log x_0 \) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} (x \log x) \)
Since both \( x \) and \( \log x \) are continuous functions in the interval \( (0, \infty) \), we can write:
\( \lim_{x \to x_0} f(x) = (\lim_{x \to x_0} x) (\lim_{x \to x_0} \log x) = x_0 \log x_0 \) Because \( \lim_{x \to x_0} f(x) = f(x_0) \), the function \( f(x) \) is continuous at all points within its domain \( (0, \infty) \).
In simple words: This function is continuous only for numbers greater than zero. This is because you cannot take the logarithm of zero or a negative number. Wherever \( \log x \) is defined, the whole function is smooth.

🎯 Exam Tip: Always pay close attention to the domain of logarithmic functions. The natural logarithm `log x` is only defined for `x > 0`, which restricts the continuity of any function involving it.

 

Question 2. Examine the continuity of the following
(vi) \( \frac{\sin x}{x^2} \)
Answer:Let \( f(x) = \frac{\sin x}{x^2} \). The function \( \sin x \) is defined and continuous for all real numbers. However, the denominator \( x^2 \) becomes zero at \( x = 0 \). A function is undefined where its denominator is zero. Therefore, \( f(x) \) is defined and continuous on the domain \( R - \{0\} \). Let \( x_0 \) be an arbitrary point in \( R - \{0\} \). First, find the value of the function at \( x_0 \):
\( f(x_0) = \frac{\sin x_0}{x_0^2} \) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} \left( \frac{\sin x}{x^2} \right) \)
Since both \( \sin x \) and \( x^2 \) are continuous, and \( x_0 \neq 0 \), we can write:
\( \lim_{x \to x_0} f(x) = \frac{\lim_{x \to x_0} \sin x}{\lim_{x \to x_0} x^2} = \frac{\sin x_0}{x_0^2} \) Because \( \lim_{x \to x_0} f(x) = f(x_0) \), the function \( f(x) \) is continuous at all points within its domain \( R - \{0\} \).
In simple words: This function is continuous everywhere except at \( x = 0 \). This is because dividing by \( x^2 \) is not allowed when \( x \) is zero, which creates a break in the graph at that point.

🎯 Exam Tip: Rational functions (fractions with polynomials) and quotients of continuous functions are discontinuous at points where the denominator is zero. Always identify these points first.

 

Question 2. Examine the continuity of the following
(vii) \( \frac{x^2 - 16}{x + 4} \)
Answer:Let \( f(x) = \frac{x^2 - 16}{x + 4} \). The function \( f(x) \) is a rational function. It is defined for all real numbers except where the denominator is zero. The denominator \( x + 4 = 0 \) when \( x = -4 \). Therefore, \( f(x) \) is defined and continuous on the domain \( R - \{-4\} \). Let \( x_0 \) be an arbitrary point in \( R - \{-4\} \). First, find the value of the function at \( x_0 \):
\( f(x_0) = \frac{x_0^2 - 16}{x_0 + 4} = \frac{(x_0 - 4)(x_0 + 4)}{x_0 + 4} = x_0 - 4 \) (since \( x_0 \neq -4 \)) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} \left( \frac{x^2 - 16}{x + 4} \right) \)
Since \( x \neq -4 \), we can simplify the expression first:
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} \left( \frac{(x - 4)(x + 4)}{x + 4} \right) = \lim_{x \to x_0} (x - 4) = x_0 - 4 \) Because \( \lim_{x \to x_0} f(x) = f(x_0) \), the function \( f(x) \) is continuous at all points within its domain \( R - \{-4\} \).
In simple words: This function is continuous everywhere except at \( x = -4 \). At \( x = -4 \), the bottom part of the fraction becomes zero, which makes the function undefined. Otherwise, the function behaves like a simple line.

🎯 Exam Tip: For rational functions, always simplify the expression first if possible to find if there's a "hole" (removable discontinuity) or a vertical asymptote. In this case, there's a hole at \( x = -4 \).

 

Question 2. Examine the continuity of the following
(viii) \( |x + 2| + |x - 1| \)
Answer:Let \( f(x) = |x + 2| + |x - 1| \). The absolute value function \( |u| \) is defined and continuous for all real numbers. Since \( f(x) \) is a sum of two absolute value functions, \( |x+2| \) and \( |x-1| \), and both are individually continuous for all real numbers, their sum \( f(x) \) will also be defined and continuous for all real numbers \( x \). Let \( x_0 \) be an arbitrary real number. First, find the value of the function at \( x_0 \):
\( f(x_0) = |x_0 + 2| + |x_0 - 1| \) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} (|x + 2| + |x - 1|) \)
Since absolute value functions are continuous, we can substitute directly:
\( \lim_{x \to x_0} f(x) = |x_0 + 2| + |x_0 - 1| \) Because \( \lim_{x \to x_0} f(x) = f(x_0) \), the function \( f(x) \) is continuous at all points in R.
In simple words: This function is continuous everywhere. Absolute value functions are always smooth, even though they have "corners" where the expression inside becomes zero. These corners don't break the graph.

🎯 Exam Tip: Functions involving absolute values like \( |ax+b| \) are continuous for all real numbers. Remember that the sum of continuous functions is also continuous.

 

Question 2. Examine the continuity of the following
(ix) \( \frac{|x - 2|}{|x + 1|} \)
Answer:Let \( f(x) = \frac{|x - 2|}{|x + 1|} \). The numerator \( |x - 2| \) is defined and continuous for all real numbers. The denominator \( |x + 1| \) is also defined and continuous for all real numbers. However, a function that is a quotient of two functions is undefined when its denominator is zero. The denominator \( |x + 1| = 0 \) when \( x + 1 = 0 \), which means \( x = -1 \). Therefore, \( f(x) \) is defined and continuous on the domain \( R - \{-1\} \). Let \( x_0 \) be an arbitrary point in \( R - \{-1\} \). First, find the value of the function at \( x_0 \):
\( f(x_0) = \frac{|x_0 - 2|}{|x_0 + 1|} \) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} \left( \frac{|x - 2|}{|x + 1|} \right) \)
Since both numerator and denominator are continuous functions and \( x_0 \neq -1 \), we can write:
\( \lim_{x \to x_0} f(x) = \frac{\lim_{x \to x_0} |x - 2|}{\lim_{x \to x_0} |x + 1|} = \frac{|x_0 - 2|}{|x_0 + 1|} \) Because \( \lim_{x \to x_0} f(x) = f(x_0) \), the function \( f(x) \) is continuous at all points within its domain \( R - \{-1\} \).
In simple words: This function is continuous everywhere except at \( x = -1 \). At this point, the bottom part of the fraction becomes zero, making the function undefined and causing a break in its graph.

🎯 Exam Tip: When dealing with fractions, always identify points where the denominator becomes zero, as these are points of discontinuity. Even absolute value functions can cause discontinuities if they are in the denominator.

 

Question 2. Examine the continuity of the following
(x) \( \cot x + \tan x \)
Answer:Let \( f(x) = \cot x + \tan x \). The function \( \cot x \) is continuous everywhere except at points where \( \sin x = 0 \), which are \( x = n\pi \) for any integer \( n \). The function \( \tan x \) is continuous everywhere except at points where \( \cos x = 0 \), which are \( x = (2n + 1)\frac{\pi}{2} \) for any integer \( n \). Therefore, \( f(x) \) is defined and continuous on the domain \( R - \{ \frac{n\pi}{2} : n \in Z \} \). This is because points like \( \pi, 2\pi, \dots \) (where \( \cot x \) is undefined) and points like \( \frac{\pi}{2}, \frac{3\pi}{2}, \dots \) (where \( \tan x \) is undefined) cover all integer multiples of \( \frac{\pi}{2} \). Let \( x_0 \) be an arbitrary point in this domain. First, find the value of the function at \( x_0 \):
\( f(x_0) = \cot x_0 + \tan x_0 \) Next, find the limit of the function as \( x \) approaches \( x_0 \):
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} (\cot x + \tan x) \)
Since both \( \cot x \) and \( \tan x \) are continuous in the chosen domain, we can write:
\( \lim_{x \to x_0} f(x) = \lim_{x \to x_0} \cot x + \lim_{x \to x_0} \tan x = \cot x_0 + \tan x_0 \) Because \( \lim_{x \to x_0} f(x) = f(x_0) \), the function \( f(x) \) is continuous at all points within its domain.
In simple words: This function has breaks at all multiples of \( \frac{\pi}{2} \) (like \( 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \), etc.). This is because \( \cot x \) has breaks at \( n\pi \) and \( \tan x \) has breaks at \( (2n+1)\frac{\pi}{2} \). The function is smooth everywhere else.

🎯 Exam Tip: When combining trigonometric functions like \( \tan x \) and \( \cot x \), be careful to identify all points where any of the individual functions are undefined. These points become discontinuities for the combined function.

 

Question 3. Find the points of discontinuity of the function \( f \), where
(i) \( f(x) = \begin{cases} 4x + 5 & \text{if } x \le 3 \\ 4x - 5 & \text{if } x > 3 \end{cases} \)
Answer:The function \( f(x) \) is defined by two different polynomial expressions. Polynomials are always continuous. So, potential discontinuity can only occur at the "joining point" where the definition changes, which is at \( x = 3 \). Let's check for continuity at \( x = 3 \): 1. **Value of the function at \( x = 3 \):**
Since \( x \le 3 \), we use the first rule:
\( f(3) = 4(3) + 5 = 12 + 5 = 17 \) 2. **Left-hand limit (LHL) at \( x = 3 \):**
As \( x \) approaches 3 from values less than 3, we use the first rule:
\( \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (4x + 5) = 4(3) + 5 = 12 + 5 = 17 \) 3. **Right-hand limit (RHL) at \( x = 3 \):**
As \( x \) approaches 3 from values greater than 3, we use the second rule:
\( \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (4x - 5) = 4(3) - 5 = 12 - 5 = 7 \) Since the Left-hand limit (17) is not equal to the Right-hand limit (7), the limit of \( f(x) \) as \( x \) approaches 3 does not exist. Therefore, the function \( f(x) \) is discontinuous at \( x = 3 \).
In simple words: This function has a break at \( x = 3 \). When you approach 3 from numbers smaller than it, the function's value goes to 17. But when you approach 3 from numbers larger than it, the value goes to 7. Since these don't match, the function jumps at \( x = 3 \).

🎯 Exam Tip: For piecewise functions, always check continuity at the points where the function's definition changes. Calculate the function value, left-hand limit, and right-hand limit, and ensure all three are equal for continuity.

 

Question 3. Find the points of discontinuity of the function \( f \), where
(ii) \( f(x) = \begin{cases} |x + 2| & \text{if } x \ge 2 \\ x^2 & \text{if } x < 2 \end{cases} \)
Answer:The function \( f(x) \) is defined by \( x^2 \) for \( x < 2 \) and \( |x+2| \) for \( x \ge 2 \). Both \( x^2 \) and \( |x+2| \) are continuous functions in their respective domains. The only point where discontinuity could occur is at the boundary, \( x = 2 \). Let's check for continuity at \( x = 2 \): 1. **Value of the function at \( x = 2 \):**
Since \( x \ge 2 \), we use the first rule:
\( f(2) = |2 + 2| = |4| = 4 \) 2. **Left-hand limit (LHL) at \( x = 2 \):**
As \( x \) approaches 2 from values less than 2, we use the second rule:
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4 \) 3. **Right-hand limit (RHL) at \( x = 2 \):**
As \( x \) approaches 2 from values greater than 2, we use the first rule:
\( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} |x + 2| = |2 + 2| = 4 \) Since \( f(2) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 4 \), the function is continuous at \( x = 2 \). Furthermore, \( f(x) = x^2 \) for \( x < 2 \) is continuous. Also, \( f(x) = |x+2| \) for \( x > 2 \) is continuous. Therefore, the function \( f(x) \) is continuous at all points in R.
In simple words: This function is continuous everywhere and has no breaks. At the point where the definition changes, \( x = 2 \), the two parts of the function meet up perfectly, with the value being 4 from both sides and at the point itself.

🎯 Exam Tip: Remember that the absolute value function \( |u| \) is continuous everywhere. For piecewise functions, carefully evaluate the limits and the function's value at the critical points where definitions change.

 

Question 3. Find the points of discontinuity of the function \( f \), where
(iii) \( f(x) = \begin{cases} x^3 - 3 & \text{if } x \le 2 \\ x^2 + 1 & \text{if } x > 2 \end{cases} \)
Answer:The function \( f(x) \) is defined by polynomial expressions. Polynomials are always continuous. So, potential discontinuity can only occur at the point where the definition changes, which is at \( x = 2 \). Let's check for continuity at \( x = 2 \): 1. **Value of the function at \( x = 2 \):**
Since \( x \le 2 \), we use the first rule:
\( f(2) = 2^3 - 3 = 8 - 3 = 5 \) 2. **Left-hand limit (LHL) at \( x = 2 \):**
As \( x \) approaches 2 from values less than 2, we use the first rule:
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^3 - 3) = 2^3 - 3 = 8 - 3 = 5 \) 3. **Right-hand limit (RHL) at \( x = 2 \):**
As \( x \) approaches 2 from values greater than 2, we use the second rule:
\( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 + 1) = 2^2 + 1 = 4 + 1 = 5 \) Since \( f(2) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 5 \), the function is continuous at \( x = 2 \). Also, \( f(x) = x^3 - 3 \) for \( x < 2 \) is continuous, and \( f(x) = x^2 + 1 \) for \( x > 2 \) is continuous. Therefore, the function \( f(x) \) is continuous at all points in R.
In simple words: This function is continuous everywhere, meaning its graph has no breaks or jumps. At the point where its definition changes, \( x = 2 \), both parts of the function meet exactly at the value of 5, making it smooth.

🎯 Exam Tip: For piecewise functions made of polynomials, the only points you need to check for discontinuity are where the definition changes. If the limits and function value match at these points, the function is continuous everywhere.

 

Question 3. Find the points of discontinuity of the function \( f \), where
(iv) \( f(x) = \begin{cases} \sin x & \text{if } 0 \le x \le \frac{\pi}{4} \\ \cos x & \text{if } \frac{\pi}{4} < x < \frac{\pi}{2} \end{cases} \)
Answer:The function \( f(x) \) is defined by \( \sin x \) for \( x \) in \( [0, \frac{\pi}{4}] \) and \( \cos x \) for \( x \) in \( (\frac{\pi}{4}, \frac{\pi}{2}) \). Both \( \sin x \) and \( \cos x \) are continuous functions. The only point where discontinuity could occur is at the boundary \( x = \frac{\pi}{4} \). Let's check for continuity at \( x = \frac{\pi}{4} \): 1. **Value of the function at \( x = \frac{\pi}{4} \):**
Since \( x \le \frac{\pi}{4} \), we use the first rule:
\( f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \) 2. **Left-hand limit (LHL) at \( x = \frac{\pi}{4} \):**
As \( x \) approaches \( \frac{\pi}{4} \) from values less than \( \frac{\pi}{4} \), we use the first rule:
\( \lim_{x \to (\frac{\pi}{4})^-} f(x) = \lim_{x \to (\frac{\pi}{4})^-} \sin x = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \) 3. **Right-hand limit (RHL) at \( x = \frac{\pi}{4} \):**
As \( x \) approaches \( \frac{\pi}{4} \) from values greater than \( \frac{\pi}{4} \), we use the second rule:
\( \lim_{x \to (\frac{\pi}{4})^+} f(x) = \lim_{x \to (\frac{\pi}{4})^+} \cos x = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \) Since \( f(\frac{\pi}{4}) = \lim_{x \to (\frac{\pi}{4})^-} f(x) = \lim_{x \to (\frac{\pi}{4})^+} f(x) = \frac{1}{\sqrt{2}} \), the function is continuous at \( x = \frac{\pi}{4} \). Also, \( f(x) = \sin x \) is continuous on \( [0, \frac{\pi}{4}] \) and \( f(x) = \cos x \) is continuous on \( (\frac{\pi}{4}, \frac{\pi}{2}) \). Therefore, the function \( f(x) \) is continuous on its entire domain \( [0, \frac{\pi}{2}) \).
In simple words: This function is continuous across its defined range. Even though it switches from \( \sin x \) to \( \cos x \) at \( x = \frac{\pi}{4} \), both parts meet exactly at the same value \( \frac{1}{\sqrt{2}} \), meaning there are no breaks.

🎯 Exam Tip: For trigonometric piecewise functions, evaluate the sine and cosine values at the boundary points. If they match, the function is continuous at that point. Trigonometric functions themselves are generally continuous unless their domain is restricted.

 

Question 4. At the given point \( x_0 \) discover whether the given function is continuous or discontinuous citing the reasons for your answer
(i) \( x_0 = 1, f(x) = \begin{cases} \frac{x^2 - 1}{x - 1} & \text{if } x \neq 1 \\ 2 & \text{if } x = 1 \end{cases} \)
Answer:We need to determine if the function \( f(x) \) is continuous at the point \( x_0 = 1 \). For a function to be continuous at a point \( x_0 \), three conditions must be met: 1. \( f(x_0) \) must be defined. 2. \( \lim_{x \to x_0} f(x) \) must exist. 3. \( \lim_{x \to x_0} f(x) = f(x_0) \). Let's check these conditions for \( x_0 = 1 \): 1. **Value of the function at \( x_0 = 1 \):**
According to the definition, when \( x = 1 \), \( f(x) = 2 \).
So, \( f(1) = 2 \). The function is defined at \( x = 1 \). 2. **Limit of the function as \( x \) approaches \( 1 \):**
For \( x \neq 1 \), we use the rule \( f(x) = \frac{x^2 - 1}{x - 1} \). We can simplify this expression:
\( \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x^2 - 1}{x - 1} \)
Since \( x^2 - 1 = (x - 1)(x + 1) \):
\( \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} \)
As \( x \to 1 \), \( x \neq 1 \), so \( (x - 1) \) is not zero and can be cancelled:
\( \lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 1) = 1 + 1 = 2 \) The limit of the function as \( x \) approaches 1 exists and is equal to 2. 3. **Compare the function value and the limit:**
We found \( f(1) = 2 \) and \( \lim_{x \to 1} f(x) = 2 \). Since \( \lim_{x \to 1} f(x) = f(1) \), all conditions for continuity are met. Therefore, the function \( f(x) \) is continuous at \( x_0 = 1 \).
In simple words: This function is continuous at \( x = 1 \). Even though the formula looks tricky, it can be simplified. The function's value at \( x=1 \) is 2, and the value it approaches from nearby points is also 2, so there's no break.

🎯 Exam Tip: For functions defined piecewise, especially rational functions with a specific value at a point of potential discontinuity, always simplify the rational part first to find the true limit. Then, compare this limit to the given function value at that point.

 

Question 5. Show that the function \( f(x) = \begin{cases} \frac { x^{3} - 1 }{ x - 1 } & \text{if } x \neq 1 \\ 3 & \text{if } x = 1 \end{cases} \) is continuous on \( (-\infty, \infty) \).
Answer: The given function is \( f(x) = \begin{cases} \frac { x^{3} - 1 }{ x - 1 } & \text{if } x \neq 1 \\ 3 & \text{if } x = 1 \end{cases} \). This function is defined for all real numbers.
We need to examine continuity at three parts of its domain: \( x < 1 \), \( x > 1 \), and \( x = 1 \).

Case (i): For \( x_0 < 1 \) (i.e., in the interval \( (-\infty, 1) \)):
For any point \( x_0 \) in this interval, \( f(x) = \frac{x^3 - 1}{x - 1} \).
The limit of \( f(x) \) as \( x \to x_0 \) is \( \lim_{x \to x_0} \frac{x^3 - 1}{x - 1} = \frac{x_0^3 - 1}{x_0 - 1} \).
The function value at \( x_0 \) is \( f(x_0) = \frac{x_0^3 - 1}{x_0 - 1} \).
Since \( \lim_{x \to x_0} f(x) = f(x_0) \), the function is continuous for all \( x_0 \in (-\infty, 1) \).

Case (ii): For \( x_0 > 1 \) (i.e., in the interval \( (1, \infty) \)):
For any point \( x_0 \) in this interval, \( f(x) = \frac{x^3 - 1}{x - 1} \).
The limit of \( f(x) \) as \( x \to x_0 \) is \( \lim_{x \to x_0} \frac{x^3 - 1}{x - 1} = \frac{x_0^3 - 1}{x_0 - 1} \).
The function value at \( x_0 \) is \( f(x_0) = \frac{x_0^3 - 1}{x_0 - 1} \).
Since \( \lim_{x \to x_0} f(x) = f(x_0) \), the function is continuous for all \( x_0 \in (1, \infty) \).

Case (iii): At \( x = 1 \):
We need to check if \( \lim_{x \to 1} f(x) = f(1) \).
From the function definition, \( f(1) = 3 \).
Now, let's find the limit as \( x \) approaches 1:
\( \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x^3 - 1}{x - 1} \)
We can factor the numerator using the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( \lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{x - 1} \)
For \( x \neq 1 \), we can cancel out \( (x - 1) \):
\( = \lim_{x \to 1} (x^2 + x + 1) \)
Now, substitute \( x = 1 \):
\( = 1^2 + 1 + 1 = 1 + 1 + 1 = 3 \).
Since \( \lim_{x \to 1} f(x) = 3 \) and \( f(1) = 3 \), the function is continuous at \( x = 1 \).

Combining all cases, the function \( f(x) \) is continuous at all points in \( (-\infty, \infty) \). This shows that the function is continuous across its entire domain.
In simple words: We checked the function in three parts: when x is less than 1, when x is greater than 1, and exactly at x=1. In all these parts, the function's value and its limit at any point were the same. This means the graph has no breaks or jumps anywhere.

🎯 Exam Tip: For piecewise functions, always check continuity at the 'changeover' points (where the definition of the function changes) and confirm that each piece is continuous in its own defined interval.

 

Question 6. For what value of \( \alpha \) is the function \( f(x) = \begin{cases} \frac { x^{4} - 1 }{ x - 1 } & \text{if } x \neq 1 \\ \alpha & \text{if } x = 1 \end{cases} \) continuous at \( x = 1 \)?
Answer: For a function to be continuous at a point, the limit of the function as it approaches that point must be equal to the function's value at that point. Here, we need \( \lim_{x \to 1} f(x) = f(1) \).

From the definition of \( f(x) \), we know that \( f(1) = \alpha \).

Next, we calculate the limit of \( f(x) \) as \( x \) approaches 1:
\( \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x^4 - 1}{x - 1} \)
We can factor the numerator \( x^4 - 1 \) using the difference of squares formula \( a^2 - b^2 = (a - b)(a + b) \):
\( x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1) \)
We can factor \( x^2 - 1 \) further as \( (x - 1)(x + 1) \).
So, \( x^4 - 1 = (x - 1)(x + 1)(x^2 + 1) \).

Now, substitute this back into the limit expression:
\( = \lim_{x \to 1} \frac{(x - 1)(x + 1)(x^2 + 1)}{x - 1} \)
Since \( x \to 1 \), \( x \neq 1 \), so we can cancel out the \( (x - 1) \) terms:
\( = \lim_{x \to 1} (x + 1)(x^2 + 1) \)
Now, substitute \( x = 1 \) into the expression:
\( = (1 + 1)(1^2 + 1) \)
\( = (2)(1 + 1) \)
\( = (2)(2) \)
\( = 4 \).

For \( f(x) \) to be continuous at \( x = 1 \), we must have \( \lim_{x \to 1} f(x) = f(1) \).
So, \( 4 = \alpha \).
Thus, the value of \( \alpha \) must be 4.
In simple words: For the function to be smooth and connected at x=1, the value it approaches must be exactly what the function is defined as at x=1. We calculated that the function approaches the number 4 as x gets close to 1. Since the function's value at x=1 is given as \( \alpha \), then \( \alpha \) must be 4.

🎯 Exam Tip: When dealing with removable discontinuities, remember that the limit of the function must exist. Factorization is a common technique to simplify expressions and find limits at points where the function is initially undefined.

 

Question 7. Graph the function. Show that \( f(x) \) continuous on \( (-\infty, \infty) \) where \( f(x) = \begin{cases} 0 & \text{if } x < 0 \\ x^2 & \text{if } 0 \le x < 2 \\ 4 & \text{if } x \ge 2 \end{cases} \).
Answer: To show that the function \( f(x) \) is continuous on \( (-\infty, \infty) \), we need to check its continuity at the points where its definition changes, which are \( x = 0 \) and \( x = 2 \). Also, we need to ensure that each piece of the function is continuous in its respective interval.

Continuity within intervals:
- For \( x < 0 \), \( f(x) = 0 \), which is a constant function and thus continuous.
- For \( 0 < x < 2 \), \( f(x) = x^2 \), which is a polynomial function and thus continuous.
- For \( x > 2 \), \( f(x) = 4 \), which is a constant function and thus continuous.

Continuity at \( x = 0 \):
1. \( f(0) = 0^2 = 0 \) (from the definition \( 0 \le x < 2 \)).
2. Left-hand limit: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 0 = 0 \).
3. Right-hand limit: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0^2 = 0 \).
Since \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0 \), the function is continuous at \( x = 0 \).

Continuity at \( x = 2 \):
1. \( f(2) = 4 \) (from the definition \( x \ge 2 \)).
2. Left-hand limit: \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4 \).
3. Right-hand limit: \( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} 4 = 4 \).
Since \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 4 \), the function is continuous at \( x = 2 \).

Since the function is continuous within each defined interval and at the transition points, it is continuous on \( (-\infty, \infty) \). This creates a smooth curve without any breaks.
In simple words: The function changes its rule at x=0 and x=2. We checked if the pieces meet smoothly at these points by looking at the limit from both sides and the function's value. Since they all matched up, the function is continuous everywhere.

🎯 Exam Tip: When graphing piecewise functions, remember to accurately represent open and closed intervals with hollow or solid circles at the transition points. Pay close attention to how the segments connect or if they create a jump.

 

Question 8. If \( f \) and \( g \) are continuous functions with \( f(3) = 5 \) and \( \lim_{x \to 3} [2f(x) - g(x)] = 4 \), find \( g(3) \).
Answer: We are given that \( f \) and \( g \) are continuous functions. This means that for any point \( c \) in their domain, \( \lim_{x \to c} f(x) = f(c) \) and \( \lim_{x \to c} g(x) = g(c) \).

We are given the following information:
1. \( f(3) = 5 \).
2. \( \lim_{x \to 3} [2f(x) - g(x)] = 4 \).

Since \( f \) is continuous at \( x = 3 \), we know that:
\( \lim_{x \to 3} f(x) = f(3) = 5 \).

Since \( g \) is continuous at \( x = 3 \), we know that:
\( \lim_{x \to 3} g(x) = g(3) \).

Now, we use the property of limits that states the limit of a sum or difference is the sum or difference of the limits, and a constant can be pulled out of the limit:
\( \lim_{x \to 3} [2f(x) - g(x)] = 2 \lim_{x \to 3} f(x) - \lim_{x \to 3} g(x) \)

Substitute the known values into this equation:
\( 4 = 2 \cdot (5) - g(3) \)
\( 4 = 10 - g(3) \)

To find \( g(3) \), rearrange the equation:
\( g(3) = 10 - 4 \)
\( g(3) = 6 \).
So, the value of \( g(3) \) is 6.
In simple words: Because both functions are continuous, their limits at x=3 are simply their values at x=3. Using the given limit equation, we can substitute the known values and easily solve for g(3).

🎯 Exam Tip: Remember that continuity means the limit equals the function value. This property is key for solving problems involving combinations of continuous functions, as it allows you to substitute function values directly for limits.

 

Question 9. Find the points at which \( f \) is discontinuous. At which of these points \( f \) is continuous from the right, from the left, or neither? Sketch the graph of \( f \).
(i) \( f(x) = \begin{cases} 2x + 1 & \text{if } x \le -1 \\ 3x & \text{if } -1 < x < 1 \\ 2x - 1 & \text{if } x \ge 1 \end{cases} \)
Answer: We need to check the continuity of \( f(x) \) at the points where its definition changes, which are \( x = -1 \) and \( x = 1 \).

Continuity at \( x = -1 \):
1. Function value: \( f(-1) = 2(-1) + 1 = -2 + 1 = -1 \).
2. Left-hand limit: \( \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (2x + 1) = 2(-1) + 1 = -1 \).
3. Right-hand limit: \( \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (3x) = 3(-1) = -3 \).
Since \( \lim_{x \to -1^-} f(x) \neq \lim_{x \to -1^+} f(x) \), the limit \( \lim_{x \to -1} f(x) \) does not exist. Therefore, \( f(x) \) is discontinuous at \( x = -1 \).
At \( x = -1 \), \( f(x) \) is continuous from the left because \( \lim_{x \to -1^-} f(x) = f(-1) \).

Continuity at \( x = 1 \):
1. Function value: \( f(1) = 2(1) - 1 = 2 - 1 = 1 \).
2. Left-hand limit: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x) = 3(1) = 3 \).
3. Right-hand limit: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 2(1) - 1 = 1 \).
Since \( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \), the limit \( \lim_{x \to 1} f(x) \) does not exist. Therefore, \( f(x) \) is discontinuous at \( x = 1 \).
At \( x = 1 \), \( f(x) \) is continuous from the right because \( \lim_{x \to 1^+} f(x) = f(1) \).

The function is discontinuous at \( x = -1 \) and \( x = 1 \). At \( x = -1 \), it is continuous from the left. At \( x = 1 \), it is continuous from the right.
In simple words: This function has breaks at x=-1 and x=1. At x=-1, if you approach from the left, the function's path connects to its value, but not from the right. At x=1, if you approach from the right, it connects, but not from the left.

🎯 Exam Tip: When graphing piecewise linear functions, plot points at the boundary of each interval. Use solid circles for inclusive endpoints and hollow circles for exclusive endpoints to correctly show continuity or discontinuity.

 

Question 9. Find the points at which \( f \) is discontinuous. At which of these points \( f \) is continuous from the right, from the left, or neither? Sketch the graph of \( f \).
(ii) \( f(x) = \begin{cases} (x - 1)^3 & \text{if } x < 0 \\ (x + 1)^3 & \text{if } x \ge 0 \end{cases} \)
Answer: We need to check the continuity of \( f(x) \) at the point where its definition changes, which is \( x = 0 \).

Continuity at \( x = 0 \):
1. Function value: \( f(0) = (0 + 1)^3 = 1^3 = 1 \) (from the definition \( x \ge 0 \)).
2. Left-hand limit: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x - 1)^3 = (0 - 1)^3 = (-1)^3 = -1 \).
3. Right-hand limit: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1)^3 = (0 + 1)^3 = 1^3 = 1 \).
Since \( \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \), the limit \( \lim_{x \to 0} f(x) \) does not exist. Therefore, \( f(x) \) is discontinuous at \( x = 0 \).
At \( x = 0 \), \( f(x) \) is continuous from the right because \( \lim_{x \to 0^+} f(x) = f(0) \).

The function is discontinuous at \( x = 0 \). At this point, it is continuous from the right.
In simple words: This function jumps at x=0. If you approach x=0 from the left, the value goes to -1. If you approach from the right, it goes to 1, which is also the function's actual value at x=0. So it's only connected from the right side.

🎯 Exam Tip: When dealing with cubic piecewise functions, evaluate points around the transition value to understand the curve's shape. Mark any jump discontinuities clearly on the graph with open and closed circles.

 

Question 10. A function \( f \) is defined as follows: Is the function \( f(x) = \begin{cases} 0 & \text{for } x < 0 \\ x & \text{for } 0 \le x < 1 \\ -x^2 + 4x - 2 & \text{for } 1 \le x < 3 \\ 4 - x & \text{for } x \ge 3 \end{cases} \) continuous?
Answer: To determine if the function \( f(x) \) is continuous, we need to check its continuity at the transition points \( x = 0 \), \( x = 1 \), and \( x = 3 \). We also confirm that each piece of the function is continuous within its defined interval (polynomials and constants are continuous).

Continuity at \( x = 0 \):
1. Function value: \( f(0) = 0 \) (from \( 0 \le x < 1 \)).
2. Left-hand limit: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 0 = 0 \).
3. Right-hand limit: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0 \).
Since \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0 \), \( f(x) \) is continuous at \( x = 0 \).

Continuity at \( x = 1 \):
1. Function value: \( f(1) = -(1)^2 + 4(1) - 2 = -1 + 4 - 2 = 1 \) (from \( 1 \le x < 3 \)).
2. Left-hand limit: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1 \).
3. Right-hand limit: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (-x^2 + 4x - 2) = -(1)^2 + 4(1) - 2 = -1 + 4 - 2 = 1 \).
Since \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 1 \), \( f(x) \) is continuous at \( x = 1 \).

Continuity at \( x = 3 \):
1. Function value: \( f(3) = 4 - 3 = 1 \) (from \( x \ge 3 \)).
2. Left-hand limit: \( \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-x^2 + 4x - 2) = -(3)^2 + 4(3) - 2 = -9 + 12 - 2 = 1 \).
3. Right-hand limit: \( \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (4 - x) = 4 - 3 = 1 \).
Since \( \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3) = 1 \), \( f(x) \) is continuous at \( x = 3 \).

Since the function is continuous at all transition points and each piece is continuous in its domain, the function is continuous everywhere on \( (-\infty, \infty) \). This makes it a smooth, unbroken function.
In simple words: We checked the function at every point where its definition changed (x=0, x=1, x=3). At all these points, the function's value and the limits from both sides matched. Since all parts connect smoothly, the entire function is continuous.

🎯 Exam Tip: When evaluating continuity for a multi-piece function, always systematically check the function value and both one-sided limits at each breakpoint. If all three are equal, the function is continuous at that point.

 

Question 11. Which of the following functions \( f \) has a removable discontinuity at \( x = x_0 \)? If the discontinuity is removable, find a function \( g \) that agrees with \( f \) for \( x \neq x_0 \) and is continuous on \( R \).
(i) \( f(x) = \frac{x^2 - 2x - 8}{x + 2} \), \( x_0 = -2 \)
Answer: The function is \( f(x) = \frac{x^2 - 2x - 8}{x + 2} \).
At \( x_0 = -2 \), the denominator becomes \( -2 + 2 = 0 \), so \( f(x) \) is not defined at \( x = -2 \). This indicates a discontinuity.

To determine if it's a removable discontinuity, we evaluate the limit of \( f(x) \) as \( x \) approaches \( -2 \):
\( \lim_{x \to -2} f(x) = \lim_{x \to -2} \frac{x^2 - 2x - 8}{x + 2} \)
Factor the numerator: \( x^2 - 2x - 8 = (x - 4)(x + 2) \).
So, the limit becomes:
\( = \lim_{x \to -2} \frac{(x - 4)(x + 2)}{x + 2} \)
For \( x \neq -2 \), we can cancel the \( (x + 2) \) terms:
\( = \lim_{x \to -2} (x - 4) \)
Now, substitute \( x = -2 \):
\( = -2 - 4 = -6 \).
Since the limit exists (it is \( -6 \)), but the function is not defined at \( x = -2 \), there is a removable discontinuity at \( x = -2 \).

To make the function continuous on \( R \), we define a new function \( g(x) \) that agrees with \( f(x) \) for \( x \neq -2 \) and fills the gap at \( x = -2 \):
\( g(x) = \begin{cases} f(x) & \text{if } x \neq -2 \\ -6 & \text{if } x = -2 \end{cases} \)
This means \( g(x) = \begin{cases} \frac{x^2 - 2x - 8}{x + 2} & \text{if } x \neq -2 \\ -6 & \text{if } x = -2 \end{cases} \)
Since \( \frac{x^2 - 2x - 8}{x + 2} = x - 4 \) for \( x \neq -2 \), the function \( g(x) \) can be simply written as \( g(x) = x - 4 \). This is a polynomial function, which is continuous for all real numbers.
In simple words: The function has a 'hole' at x=-2 because it's undefined there, but as you get very close to -2, the function approaches -6. So, we can 'fill the hole' by defining the function as -6 at x=-2, making it smooth everywhere.

🎯 Exam Tip: A discontinuity is removable if the limit exists at that point but the function is either undefined or has a different value. Factoring the numerator and denominator is a key step to simplify and find the limit.

 

Question 11. Which of the following functions \( f \) has a removable discontinuity at \( x = x_0 \)? If the discontinuity is removable, find a function \( g \) that agrees with \( f \) for \( x \neq x_0 \) and is continuous on \( R \).
(ii) \( f(x) = \frac{x^3 + 64}{x + 4} \), \( x_0 = -4 \)
Answer: The function is \( f(x) = \frac{x^3 + 64}{x + 4} \).
At \( x_0 = -4 \), the denominator becomes \( -4 + 4 = 0 \), so \( f(x) \) is not defined at \( x = -4 \). This indicates a discontinuity.

To determine if it's a removable discontinuity, we evaluate the limit of \( f(x) \) as \( x \) approaches \( -4 \):
\( \lim_{x \to -4} f(x) = \lim_{x \to -4} \frac{x^3 + 64}{x + 4} \)
Factor the numerator using the sum of cubes formula: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Here, \( a = x \) and \( b = 4 \), so \( x^3 + 64 = x^3 + 4^3 = (x + 4)(x^2 - 4x + 16) \).
So, the limit becomes:
\( = \lim_{x \to -4} \frac{(x + 4)(x^2 - 4x + 16)}{x + 4} \)
For \( x \neq -4 \), we can cancel the \( (x + 4) \) terms:
\( = \lim_{x \to -4} (x^2 - 4x + 16) \)
Now, substitute \( x = -4 \):
\( = (-4)^2 - 4(-4) + 16 \)
\( = 16 + 16 + 16 \)
\( = 48 \).
Since the limit exists (it is \( 48 \)), but the function is not defined at \( x = -4 \), there is a removable discontinuity at \( x = -4 \).

To make the function continuous on \( R \), we define a new function \( g(x) \) that agrees with \( f(x) \) for \( x \neq -4 \) and fills the gap at \( x = -4 \):
\( g(x) = \begin{cases} f(x) & \text{if } x \neq -4 \\ 48 & \text{if } x = -4 \end{cases} \)
This means \( g(x) = \begin{cases} \frac{x^3 + 64}{x + 4} & \text{if } x \neq -4 \\ 48 & \text{if } x = -4 \end{cases} \)
Since \( \frac{x^3 + 64}{x + 4} = x^2 - 4x + 16 \) for \( x \neq -4 \), the function \( g(x) \) can be simply written as \( g(x) = x^2 - 4x + 16 \). This is a polynomial function, which is continuous for all real numbers.
In simple words: The function has a missing point at x=-4 because its denominator becomes zero. However, we found that as x gets very close to -4, the function's value gets very close to 48. So, we can create a new function that is exactly 48 at x=-4 and the same as the original function everywhere else, making it continuous.

🎯 Exam Tip: For removable discontinuities, look for opportunities to simplify the function algebraically, often by factoring. The value of the limit after simplification indicates what the function's value should be at the point of discontinuity for it to be continuous.

 

Question 11. (iii) Which of the following functions \( f \) has a removable discontinuity at \( x = x_0 \)? If the discontinuity is removable, find a function \( g \) that agrees with \( f \) for \( x \ne x_0 \) and is continuous on R for \( f(x) = \frac { 3 - \sqrt { x } }{ 9 - x } \), \( x_0 = 9 \).
Answer: The given function is \( f(x) = \frac { 3 - \sqrt { x } }{ 9 - x } \). This function is not defined at \( x = 9 \). To check if it has a removable discontinuity at \( x = 9 \), we find the limit of \( f(x) \) as \( x \to 9 \):
\( \lim_{x \to 9} f(x) = \lim_{x \to 9} \frac { 3 - \sqrt { x } }{ 9 - x } \)
We can rewrite the denominator as \( 9 - x = 3^2 - (\sqrt{x})^2 = (3 - \sqrt{x})(3 + \sqrt{x}) \).
So, \( \lim_{x \to 9} \frac { 3 - \sqrt { x } }{ (3 - \sqrt { x })(3 + \sqrt { x }) } \)
Since \( x \to 9 \), \( x \ne 9 \), so \( 3 - \sqrt{x} \ne 0 \). We can cancel out the common factor:
\( \lim_{x \to 9} \frac { 1 }{ 3 + \sqrt { x } } \)
Now, substitute \( x = 9 \):
\( = \frac { 1 }{ 3 + \sqrt { 9 } } = \frac { 1 }{ 3 + 3 } = \frac { 1 }{ 6 } \)
Since the limit exists, the function has a removable discontinuity at \( x = 9 \). We can redefine the function to be continuous on R by setting \( g(x) \) equal to the limit at \( x = 9 \).
The new function \( g(x) \) that agrees with \( f(x) \) for \( x \ne 9 \) and is continuous on R is:
\[ g(x) = \begin{cases} \frac { 3 - \sqrt { x } }{ 9 - x } & \text{if } x \ne 9 \\ \frac { 1 }{ 6 } & \text{if } x = 9 \end{cases} \]
This new function \( g(x) \) is defined at all points of R and is continuous. This process fills the 'hole' in the function's graph at \( x=9 \).
In simple words: The original function was undefined at \( x = 9 \). By calculating what value the function approaches as \( x \) gets close to \( 9 \), which is \( \frac{1}{6} \), we can make the function continuous. We create a new function that is the same as the old one everywhere else, but at \( x = 9 \), its value is \( \frac{1}{6} \).

🎯 Exam Tip: To remove a discontinuity, always calculate the limit of the function at the point of discontinuity. If the limit exists, define the function at that point to be equal to the limit value.

 

Question 12. Find the constant b that makes g continuous on \( (-\infty, \infty) \). \( g(x) = \begin{cases} x^2 - b^2 & \text{if } x < 4 \\ bx + 20 & \text{if } x \ge 4 \end{cases} \)
Answer: For the function \( g(x) \) to be continuous on \( (-\infty, \infty) \), it must be continuous at every point. Since both parts of the function (polynomials) are continuous on their respective intervals, we only need to ensure continuity at the boundary point, \( x = 4 \).
For \( g(x) \) to be continuous at \( x = 4 \), the left-hand limit, the right-hand limit, and the function's value at \( x = 4 \) must all be equal.
Left-hand limit: \( \lim_{x \to 4^-} g(x) = \lim_{x \to 4^-} (x^2 - b^2) = 4^2 - b^2 = 16 - b^2 \)
Right-hand limit: \( \lim_{x \to 4^+} g(x) = \lim_{x \to 4^+} (bx + 20) = b(4) + 20 = 4b + 20 \)
Function value at \( x = 4 \): \( g(4) = b(4) + 20 = 4b + 20 \)
For continuity, we must have: Left-hand limit = Right-hand limit
\( 16 - b^2 = 4b + 20 \)
Now, we solve this quadratic equation for \( b \):
\( 0 = b^2 + 4b + 20 - 16 \)
\( 0 = b^2 + 4b + 4 \)
We recognize this as a perfect square:
\( (b + 2)^2 = 0 \)
Taking the square root of both sides gives:
\( b + 2 = 0 \)
\( b = -2 \)
Therefore, the value of the constant \( b \) that makes the function \( g \) continuous everywhere is \( -2 \). This ensures the two pieces of the function meet smoothly at \( x=4 \).
In simple words: For the function to be smooth everywhere, the two different rules for \( g(x) \) must give the same answer when \( x \) is exactly \( 4 \). We set the result of the first rule for \( x < 4 \) equal to the result of the second rule for \( x \ge 4 \) at \( x=4 \). Solving this equation gives us the value of \( b \) as \( -2 \).

🎯 Exam Tip: For piecewise functions to be continuous, always ensure that the left-hand limit and the right-hand limit at the critical points (where the function definition changes) are equal to the function's value at that point.

 

Question 13. Consider the function \( f (x) = x \sin \frac{\pi}{x} \). What value must we give \( f (0) \) in order to make the function continuous everywhere?
Answer: The function is given by \( f(x) = x \sin \frac{\pi}{x} \). This function is not defined at \( x = 0 \) because of the \( \frac{\pi}{x} \) term. To make the function continuous everywhere, we need to define \( f(0) \) such that it equals the limit of \( f(x) \) as \( x \to 0 \).
We need to find \( \lim_{x \to 0} x \sin \frac{\pi}{x} \).
We know that the sine function is bounded, meaning \( -1 \le \sin \frac{\pi}{x} \le 1 \) for all \( x \ne 0 \).
If we multiply the inequality by \( |x| \) (since \( x \) can be negative, we use absolute value for proper inequality scaling):
\( -|x| \le x \sin \frac{\pi}{x} \le |x| \)
As \( x \to 0 \), we know that \( |x| \to 0 \).
By the Squeeze Theorem, since \( \lim_{x \to 0} (-|x|) = 0 \) and \( \lim_{x \to 0} |x| = 0 \), it follows that:
\( \lim_{x \to 0} x \sin \frac{\pi}{x} = 0 \)
Therefore, to make the function continuous everywhere, we must define \( f(0) = 0 \). This fills the hole at \( x=0 \) and makes the function smooth. The Squeeze Theorem is a powerful tool for limits of oscillatory functions.
In simple words: The function is tricky at \( x=0 \) because \( \frac{\pi}{x} \) is undefined. However, because the sine part always stays between \( -1 \) and \( 1 \), and it is multiplied by \( x \), as \( x \) gets closer to \( 0 \), the whole expression \( x \sin \frac{\pi}{x} \) also gets closer to \( 0 \). So, to make the function continuous, we should set \( f(0) \) to be \( 0 \).

🎯 Exam Tip: For limits involving \( x \sin(\frac{1}{x}) \) or similar forms as \( x \to 0 \), remember to apply the Squeeze Theorem. The key is that \( \sin(\text{anything}) \) is always between \( -1 \) and \( 1 \).

 

Question 14. The function \( f(x) = \frac{x^2 - 1}{x^3 - 1} \) is not defined at \( x = 1 \). What value must we give \( f(1) \) in order to make \( f(x) \) continuous at \( x = 1 \)?
Answer: The given function is \( f(x) = \frac{x^2 - 1}{x^3 - 1} \). It is not defined at \( x = 1 \) because substituting \( x = 1 \) results in \( \frac{0}{0} \), which is an indeterminate form. To make the function continuous at \( x = 1 \), we must define \( f(1) \) to be equal to the limit of \( f(x) \) as \( x \to 1 \).
First, we factorize the numerator and the denominator:
Numerator: \( x^2 - 1 = (x - 1)(x + 1) \)
Denominator: \( x^3 - 1 = (x - 1)(x^2 + x + 1) \) (using the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \))
Now, we find the limit:
\( \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{(x - 1)(x + 1)}{(x - 1)(x^2 + x + 1)} \)
Since \( x \to 1 \), \( x \ne 1 \), so \( x - 1 \ne 0 \). We can cancel the common factor \( (x - 1) \):
\( = \lim_{x \to 1} \frac{x + 1}{x^2 + x + 1} \)
Now, substitute \( x = 1 \) into the simplified expression:
\( = \frac{1 + 1}{1^2 + 1 + 1} = \frac{2}{1 + 1 + 1} = \frac{2}{3} \)
Therefore, to make the function \( f(x) \) continuous at \( x = 1 \), we must define \( f(1) = \frac{2}{3} \). This is a common method for handling removable discontinuities.
In simple words: The function doesn't have a value at \( x=1 \). We need to find what value it "should" have to become smooth. We do this by simplifying the fraction and then putting \( x=1 \) into the simplified version. This gives us \( \frac{2}{3} \), which is the value \( f(1) \) needs to be for continuity.

🎯 Exam Tip: When faced with an indeterminate form like \( \frac{0}{0} \) in a limit problem, always try to factorize the numerator and denominator to cancel common factors before substituting the limit value.

 

Question 15. State how continuity is destroyed at \( x = x_0 \) for each of the following graphs.
Answer:
(a) For the given graph (a):

The continuity is destroyed because the left-hand limit and the right-hand limit at \( x = x_0 \) do not coincide. This is a type of jump discontinuity.
(b) For the given graph (b):

The continuity is destroyed because the function \( f(x) \) is not defined at \( x = x_0 \). There is a 'hole' in the graph at this point, but the limit exists.
(c) For the given graph (c):

The continuity is destroyed because the limit of \( f(x) \) does not exist at \( x = x_0 \). The function approaches infinity, indicating a vertical asymptote.
(d) For the given graph (d):

The continuity is destroyed because the left-hand limit and the right-hand limit at \( x = x_0 \) do not coincide. This is another example of a jump discontinuity.
In simple words: For a function to be continuous, its graph should be a single, unbroken line. In these cases, the graph is broken at \( x_0 \) because: (a) the line jumps, meaning the left and right sides don't meet; (b) there's a hole in the line, so the function has no value there; (c) the line goes off to infinity, so the limit doesn't exist; and (d) similar to (a), the line jumps to a different level.

🎯 Exam Tip: Remember the three conditions for continuity at a point \( x_0 \): (1) \( f(x_0) \) must exist, (2) \( \lim_{x \to x_0} f(x) \) must exist, and (3) \( \lim_{x \to x_0} f(x) = f(x_0) \). If any of these conditions fail, the function is discontinuous.