Samacheer Kalvi Class 11 Maths Solutions Chapter 9 Limits and Continuity Exercise 9.1

In Problems 1 – 6, complete the table using calculator and use the result to estimate the limit.

 

Question 1. \( \lim_{x \to 2} \frac{x-2}{x^2 - x - 2} \)

Answer:The function is \( f(x) = \frac{x-2}{x^2 - x - 2} \). We can simplify the denominator by factoring it. \( f(x) = \frac{x-2}{(x-2)(x+1)} \) As \( x \) approaches 2, but \( x \neq 2 \), we can cancel out the common term \( (x-2) \). So, \( f(x) = \frac{1}{x+1} \) for \( x \neq 2 \). Now, let's complete the table by substituting the given \( x \) values into the simplified function:As \( x \) approaches 2 from both sides, the value of \( f(x) \) approaches \( \frac{1}{3} \) or approximately 0.33. This means the limit exists and equals 0.33.
\( \implies \)\( \lim_{x \to 2} \frac{x-2}{x^2 - x - 2} = \lim_{x \to 2} \frac{1}{x+1} = \frac{1}{2+1} = \frac{1}{3} \approx 0.33 \)In simple words: First, simplify the fraction by removing the common part from top and bottom. Then, put the number 2 into the simplified fraction. The result is what the function gets closer and closer to.

🎯 Exam Tip: When dealing with limits, always try to simplify the expression by factoring and canceling common terms before direct substitution, especially if direct substitution leads to an undefined form like \( \frac{0}{0} \).

 

Question 2. \( \lim_{x \to 2} \frac{x-2}{x^2 - 4} \)

Answer:The function is \( f(x) = \frac{x-2}{x^2 - 4} \). We can simplify the denominator using the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \). \( f(x) = \frac{x-2}{(x-2)(x+2)} \) As \( x \) approaches 2, but \( x \neq 2 \), we can cancel out the common term \( (x-2) \). So, \( f(x) = \frac{1}{x+2} \) for \( x \neq 2 \). Now, let's complete the table by substituting the given \( x \) values into the simplified function:As \( x \) approaches 2 from both sides, the value of \( f(x) \) approaches \( \frac{1}{4} \) or 0.25. This shows that the limit exists.
\( \implies \)\( \lim_{x \to 2} \frac{x-2}{x^2 - 4} = \lim_{x \to 2} \frac{1}{x+2} = \frac{1}{2+2} = \frac{1}{4} = 0.25 \)In simple words: Break down the bottom part of the fraction using the rule for differences of squares. This helps you simplify the fraction. Then, put 2 in place of \( x \) to find what the function approaches.

🎯 Exam Tip: Remember to simplify expressions by factoring before evaluating limits, especially for indeterminate forms. The difference of squares formula \( (a^2-b^2) = (a-b)(a+b) \) is very useful here.

 

Question 3. \( \lim_{x \to 0} \frac{\sqrt{x+3} - \sqrt{3}}{x} \)

Answer:Let \( f(x) = \frac{\sqrt{x+3} - \sqrt{3}}{x} \). If we substitute \( x=0 \), we get \( \frac{0}{0} \), which is an indeterminate form. To solve this, we can multiply the numerator and denominator by the conjugate of the numerator. The conjugate of \( \sqrt{x+3} - \sqrt{3} \) is \( \sqrt{x+3} + \sqrt{3} \). \( f(x) = \frac{\sqrt{x+3} - \sqrt{3}}{x} \times \frac{\sqrt{x+3} + \sqrt{3}}{\sqrt{x+3} + \sqrt{3}} \)
\( \implies \)\( f(x) = \frac{(x+3) - 3}{x(\sqrt{x+3} + \sqrt{3})} \)
\( \implies \)\( f(x) = \frac{x}{x(\sqrt{x+3} + \sqrt{3})} \) For \( x \neq 0 \), we can cancel \( x \) from the numerator and denominator.
\( \implies \)\( f(x) = \frac{1}{\sqrt{x+3} + \sqrt{3}} \) Now, let's complete the table:As \( x \) approaches 0 from both sides, the value of \( f(x) \) approaches \( \frac{1}{2\sqrt{3}} \) or approximately 0.288.
\( \implies \)\( \lim_{x \to 0} \frac{\sqrt{x+3} - \sqrt{3}}{x} = \lim_{x \to 0} \frac{1}{\sqrt{x+3} + \sqrt{3}} = \frac{1}{\sqrt{0+3} + \sqrt{3}} = \frac{1}{2\sqrt{3}} \) \( \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{6} \approx \frac{1.732}{6} \approx 0.288 \)In simple words: When you have square roots in a limit problem that gives \( \frac{0}{0} \), multiply the top and bottom by the "opposite sign" version of the square root part. This helps to remove the square roots from the top and lets you simplify the fraction to find the limit.

🎯 Exam Tip: For limits involving square roots that result in an indeterminate form, rationalizing the numerator (or denominator) by multiplying by the conjugate is a standard technique. Remember that \( (a-b)(a+b) = a^2 - b^2 \).

 

Question 4. \( \lim_{x \to -3} \frac{\sqrt{1-x} - 2}{x+3} \)

Answer:Let \( f(x) = \frac{\sqrt{1-x} - 2}{x+3} \). If we substitute \( x=-3 \), we get \( \frac{\sqrt{1-(-3)}-2}{-3+3} = \frac{\sqrt{4}-2}{0} = \frac{0}{0} \), which is an indeterminate form. We need to multiply the numerator and denominator by the conjugate of the numerator. The conjugate of \( \sqrt{1-x} - 2 \) is \( \sqrt{1-x} + 2 \). \( f(x) = \frac{\sqrt{1-x} - 2}{x+3} \times \frac{\sqrt{1-x} + 2}{\sqrt{1-x} + 2} \)
\( \implies \)\( f(x) = \frac{(1-x) - 2^2}{(x+3)(\sqrt{1-x} + 2)} \)
\( \implies \)\( f(x) = \frac{1-x-4}{(x+3)(\sqrt{1-x} + 2)} \)
\( \implies \)\( f(x) = \frac{-x-3}{(x+3)(\sqrt{1-x} + 2)} \)
\( \implies \)\( f(x) = \frac{-(x+3)}{(x+3)(\sqrt{1-x} + 2)} \) For \( x \neq -3 \), we can cancel \( (x+3) \) from the numerator and denominator.
\( \implies \)\( f(x) = \frac{-1}{\sqrt{1-x} + 2} \) Now, let's complete the table:As \( x \) approaches -3 from both sides, the value of \( f(x) \) approaches \( \frac{-1}{4} \) or -0.25. This shows that the limit exists.
\( \implies \)\( \lim_{x \to -3} \frac{\sqrt{1-x} - 2}{x+3} = \lim_{x \to -3} \frac{-1}{\sqrt{1-x} + 2} = \frac{-1}{\sqrt{1-(-3)} + 2} = \frac{-1}{\sqrt{4} + 2} = \frac{-1}{2+2} = \frac{-1}{4} = -0.25 \)In simple words: When you see square roots that make the fraction unclear, multiply the top and bottom by the "conjugate" (the same numbers but with the opposite sign in the middle). This step removes the square root from the top and helps simplify the fraction so you can find the limit.

🎯 Exam Tip: Be careful with signs when substituting negative values into the expression, especially under the square root. Also, remember that squaring \( \sqrt{1-x} \) gives \( 1-x \), and squaring 2 gives 4, so \( (\sqrt{1-x}-2)(\sqrt{1-x}+2) = (1-x) - 4 \).

 

Question 5. \( \lim_{x \to 0} \frac{\sin x}{x} \)

Answer:Let \( f(x) = \frac{\sin x}{x} \). When \( x=0 \), this gives \( \frac{\sin 0}{0} = \frac{0}{0} \), which is an indeterminate form. We need to evaluate the limit using values close to 0. It is important to use radians for \( x \) when calculating \( \sin x \) for this limit.As \( x \) approaches 0 from both sides, the value of \( f(x) \) approaches 1. This shows that the limit exists.
\( \implies \)\( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) This is a very important standard limit in calculus.In simple words: As \( x \) gets very, very close to zero, the value of \( \sin x \) divided by \( x \) gets very, very close to 1. Always remember to use radians when you calculate \( \sin x \) for this type of limit problem.

🎯 Exam Tip: This is a fundamental trigonometric limit. Make sure to memorize that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Also, confirm your calculator is in radian mode for such problems.

 

Question 6. \( \lim_{x \to 0} \frac{\cos x - 1}{x} \)

Answer:Let \( f(x) = \frac{\cos x - 1}{x} \). When \( x=0 \), this gives \( \frac{\cos 0 - 1}{0} = \frac{1-1}{0} = \frac{0}{0} \), which is an indeterminate form. We need to evaluate the limit by looking at values close to 0. It is important to use radians for \( x \) when calculating \( \cos x \).Looking at the values in the table, as \( x \) approaches 0 from both positive and negative sides, \( f(x) \) approaches 0. This is another important standard limit.
\( \implies \)\( \lim_{x \to 0} \frac{\cos x - 1}{x} = 0 \)In simple words: When \( x \) gets very, very close to zero, the value of \( \cos x \) minus 1, all divided by \( x \), gets very close to 0. This limit is often used to figure out how fast \( \cos x \) changes near zero.

🎯 Exam Tip: This is another key trigonometric limit to remember. Ensure your calculator is in radian mode for accurate calculations when evaluating trigonometric functions for limits.

 

Question 7. Evaluate \( \lim_{x \to 3} (4 - x) \) using the graph.

Answer:The function is \( f(x) = 4 - x \). This is a linear function, which means its graph is a straight line. From the provided graph, we can see that as \( x \) approaches 3 from the left side (values like 2.9, 2.99), the value of \( f(x) \) approaches \( 4 - 3 = 1 \). Similarly, as \( x \) approaches 3 from the right side (values like 3.1, 3.01), the value of \( f(x) \) also approaches \( 4 - 3 = 1 \). Since the function approaches the same value from both sides, the limit exists and is equal to 1. The value of the function at \( x=3 \) is \( f(3) = 4-3=1 \). Therefore, \( \lim_{x \to 3} (4 - x) = 1 \).In simple words: When you look at the graph of \( y = 4 - x \) and move your finger along the line closer and closer to where \( x \) is 3, you will see that the \( y \) value gets closer and closer to 1. So, the limit is 1.

🎯 Exam Tip: For continuous functions (like polynomials or linear functions), the limit as \( x \) approaches a point \( a \) is simply the value of the function at \( a \), i.e., \( \lim_{x \to a} f(x) = f(a) \).

 

Question 8. Evaluate \( \lim_{x \to 1} (x^2 + 2) \) using the graph.

Answer:The function is \( f(x) = x^2 + 2 \). This is a quadratic function, which means its graph is a parabola. From the provided graph, we can see that as \( x \) approaches 1 from the left side (values like 0.9, 0.99), the value of \( f(x) \) approaches \( 1^2 + 2 = 3 \). Similarly, as \( x \) approaches 1 from the right side (values like 1.1, 1.01), the value of \( f(x) \) also approaches \( 1^2 + 2 = 3 \). Since the function approaches the same value from both sides, the limit exists and is equal to 3. The value of the function at \( x=1 \) is \( f(1) = 1^2+2 = 3 \). Therefore, \( \lim_{x \to 1} (x^2 + 2) = 3 \).In simple words: If you trace the graph of \( y = x^2 + 2 \) and get closer and closer to \( x \) being 1, you will see that the \( y \) value gets closer and closer to 3. So, the limit is 3.

🎯 Exam Tip: Quadratic functions are also continuous. For any continuous function, the limit as \( x \) approaches \( a \) is simply \( f(a) \). Graphically, this means there are no jumps or holes at that point.

 

Question 9. Evaluate \( \lim_{x \to 2} f(x) \) where \( f(x) = \begin{cases} 4 - x, & x \neq 2 \\ 0, & x = 2 \end{cases} \)

Answer:The function is defined as \( f(x) = 4 - x \) when \( x \neq 2 \), and \( f(x) = 0 \) when \( x = 2 \). To find the limit as \( x \) approaches 2, we only care about the values of \( f(x) \) when \( x \) is close to 2, but not exactly 2. So, we use the rule \( f(x) = 4 - x \) for finding the limit. As \( x \) approaches 2 from the left (e.g., 1.9, 1.99): \( f(x) = 4 - x \) approaches \( 4 - 2 = 2 \). As \( x \) approaches 2 from the right (e.g., 2.1, 2.01): \( f(x) = 4 - x \) approaches \( 4 - 2 = 2 \). Since the left-hand limit and the right-hand limit both approach 2, the limit exists and is 2. Note that the actual value of the function at \( x=2 \) is \( f(2)=0 \), which is different from the limit. This just means the function has a "hole" at \( x=2 \) and is defined differently at that exact point. Therefore, \( \lim_{x \to 2} f(x) = 2 \).In simple words: When calculating a limit, we look at what the function values are getting close to, not what the function actually is at that exact point. Even though \( f(2) \) is 0, the numbers around \( x=2 \) (like 1.99 or 2.01) lead \( f(x) \) to be close to 2. So the limit is 2.

🎯 Exam Tip: The existence of a limit at a point does not depend on the function's value at that specific point. It only depends on the values of the function near that point. A hole in the graph at the limit point is common.

 

Question 10. Evaluate \( \lim_{x \to 1} f(x) \) where \( f(x) = \begin{cases} x^2 + 2, & x \neq 1 \\ 1, & x = 1 \end{cases} \)

Answer:The function is defined as \( f(x) = x^2 + 2 \) when \( x \neq 1 \), and \( f(x) = 1 \) when \( x = 1 \). To find the limit as \( x \) approaches 1, we use the rule for \( x \neq 1 \), which is \( f(x) = x^2 + 2 \). As \( x \) approaches 1 from the left (e.g., 0.9, 0.99): \( f(x) = x^2 + 2 \) approaches \( 1^2 + 2 = 3 \). As \( x \) approaches 1 from the right (e.g., 1.1, 1.01): \( f(x) = x^2 + 2 \) approaches \( 1^2 + 2 = 3 \). Since the left-hand limit and the right-hand limit both approach 3, the limit exists and is 3. The actual value of the function at \( x=1 \) is \( f(1)=1 \), which is different from the limit. This indicates a removable discontinuity, or a "hole" in the graph at \( x=1 \) that has been filled with a different point. Therefore, \( \lim_{x \to 1} f(x) = 3 \).In simple words: To find the limit, we look at the part of the function that applies when \( x \) is very close to 1 but not exactly 1. For values near 1, \( x^2 + 2 \) becomes 3. So, the limit is 3, even though the function itself is set to 1 exactly at \( x=1 \).

🎯 Exam Tip: Always evaluate the limit of a piecewise function using the part of the definition that applies to values *near* the limit point, not necessarily the definition *at* the limit point itself. A difference between \( \lim f(x) \) and \( f(a) \) indicates a point of discontinuity.

 

Question 11. To find \( \lim_{x \to 3} \frac{1}{x-3} \) using the graph.

Answer:The function is \( f(x) = \frac{1}{x-3} \). From the graph of \( y = \frac{1}{x-3} \), we can observe its behavior around \( x=3 \). As \( x \) approaches 3 from the left side (values like 2.9, 2.99), the denominator \( x-3 \) becomes a small negative number. This makes \( f(x) \) approach \( -\infty \).
\( \implies \)\( \lim_{x \to 3^-} \frac{1}{x-3} = -\infty \) As \( x \) approaches 3 from the right side (values like 3.1, 3.01), the denominator \( x-3 \) becomes a small positive number. This makes \( f(x) \) approach \( +\infty \).
\( \implies \)\( \lim_{x \to 3^+} \frac{1}{x-3} = +\infty \) Since the left-hand limit \( (-\infty) \) and the right-hand limit \( (+\infty) \) are not equal, the limit of the function as \( x \) approaches 3 does not exist. The graph shows a vertical asymptote at \( x=3 \). Therefore, \( \lim_{x \to 3} \frac{1}{x-3} \) does not exist at \( x = 3 \).In simple words: If you look at the graph, as \( x \) gets close to 3 from the left, the line goes down forever. As \( x \) gets close to 3 from the right, the line goes up forever. Because it goes in different directions, there is no single limit at \( x=3 \).

🎯 Exam Tip: If the left-hand limit and the right-hand limit approach different values (including positive or negative infinity), the overall limit does not exist. This typically occurs at vertical asymptotes for rational functions.

 

Question 12. Evaluate \( \lim_{x \to 5} \frac{|x-5|}{x-5} \) using the graph.

Answer:The function is \( f(x) = \frac{|x-5|}{x-5} \). This is a piecewise function based on the definition of the absolute value. If \( x-5 > 0 \) (i.e., \( x > 5 \)), then \( |x-5| = x-5 \). So, \( f(x) = \frac{x-5}{x-5} = 1 \) for \( x > 5 \). If \( x-5 < 0 \) (i.e., \( x < 5 \)), then \( |x-5| = -(x-5) \). So, \( f(x) = \frac{-(x-5)}{x-5} = -1 \) for \( x < 5 \). At \( x=5 \), the function is undefined because the denominator becomes 0. Now let's find the left-hand limit and right-hand limit as \( x \) approaches 5: As \( x \to 5^- \) (from the left, \( x < 5 \)), \( f(x) = -1 \). So, \( \lim_{x \to 5^-} f(x) = -1 \). As \( x \to 5^+ \) (from the right, \( x > 5 \)), \( f(x) = 1 \). So, \( \lim_{x \to 5^+} f(x) = 1 \). Since the left-hand limit \( (-1) \) and the right-hand limit \( (1) \) are not equal, the limit of the function as \( x \) approaches 5 does not exist. The graph shows a jump discontinuity at \( x=5 \). Therefore, \( \lim_{x \to 5} \frac{|x-5|}{x-5} \) does not exist at \( x = 5 \).In simple words: The function acts differently depending on whether \( x \) is smaller or larger than 5. When \( x \) is less than 5, the function value is always -1. When \( x \) is greater than 5, the function value is always 1. Because it jumps from -1 to 1 at \( x=5 \), there is no single limit.

🎯 Exam Tip: Functions involving absolute values often have different left-hand and right-hand limits, leading to the overall limit not existing. Always analyze the function's definition for \( x < a \) and \( x > a \) separately.

 

Question 13. Evaluate \( \lim_{x \to 1} \sin (\pi x) \) using the graph.

Answer:The function is \( f(x) = \sin(\pi x) \). This is a continuous trigonometric function. From the provided graph of \( y = \sin(\pi x) \), we can observe its behavior as \( x \) approaches 1. As \( x \) approaches 1 from the left side (e.g., 0.9, 0.99), the value of \( \pi x \) approaches \( \pi \), and \( \sin(\pi x) \) approaches \( \sin(\pi) = 0 \). As \( x \) approaches 1 from the right side (e.g., 1.1, 1.01), the value of \( \pi x \) also approaches \( \pi \), and \( \sin(\pi x) \) approaches \( \sin(\pi) = 0 \). Since the function approaches the same value from both sides, the limit exists and is equal to 0. The value of the function at \( x=1 \) is \( f(1) = \sin(\pi \cdot 1) = \sin(\pi) = 0 \). Therefore, \( \lim_{x \to 1} \sin(\pi x) = 0 \).In simple words: When you look at the graph of \( y = \sin(\pi x) \), as \( x \) gets closer and closer to 1, the curve goes towards the point where \( y \) is 0. So, the limit is 0.

🎯 Exam Tip: For continuous functions, direct substitution can be used to find the limit. Trigonometric functions like \( \sin x \) and \( \cos x \) are continuous everywhere, so \( \lim_{x \to a} \sin(kx) = \sin(ka) \).

 

Question 14. Evaluate \( \lim_{x \to 0} \sec x \) using the graph.

Answer:The function is \( f(x) = \sec x \), which is defined as \( \frac{1}{\cos x} \). The graph provided shows the curve of \( y = \sec x \). As \( x \) approaches 0 from the left side (e.g., -0.1, -0.01), \( \cos x \) approaches \( \cos 0 = 1 \), so \( \sec x \) approaches \( \frac{1}{1} = 1 \). As \( x \) approaches 0 from the right side (e.g., 0.1, 0.01), \( \cos x \) approaches \( \cos 0 = 1 \), so \( \sec x \) approaches \( \frac{1}{1} = 1 \). Since the function approaches the same value from both sides, the limit exists and is equal to 1. The value of the function at \( x=0 \) is \( f(0) = \sec 0 = \frac{1}{\cos 0} = \frac{1}{1} = 1 \). Therefore, \( \lim_{x \to 0} \sec x = 1 \).In simple words: When you look at the graph of \( y = \sec x \) and move your finger closer and closer to where \( x \) is 0, the curve goes towards the point where \( y \) is 1. So, the limit is 1.

🎯 Exam Tip: The secant function \( \sec x \) is continuous wherever \( \cos x \neq 0 \). Since \( \cos 0 = 1 \), there are no issues with continuity at \( x=0 \), allowing for direct substitution to find the limit.

 

Question 15. Evaluate \( \lim_{x \to \frac{\pi}{2}} \tan x \) using the graph.

Answer:The function is \( f(x) = \tan x \). This function is defined as \( \frac{\sin x}{\cos x} \). The graph provided shows the curve of \( y = \tan x \). The function \( \tan x \) has vertical asymptotes where \( \cos x = 0 \). This occurs at \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \) (and negative odd multiples of \( \frac{\pi}{2} \)). As \( x \) approaches \( \frac{\pi}{2} \) from the left side (values slightly less than \( \frac{\pi}{2} \)), \( \sin x \) approaches 1 and \( \cos x \) approaches 0 from the positive side. Thus, \( \tan x \) approaches \( +\infty \).
\( \implies \)\( \lim_{x \to (\frac{\pi}{2})^-} \tan x = +\infty \) As \( x \) approaches \( \frac{\pi}{2} \) from the right side (values slightly greater than \( \frac{\pi}{2} \)), \( \sin x \) approaches 1 and \( \cos x \) approaches 0 from the negative side. Thus, \( \tan x \) approaches \( -\infty \).
\( \implies \)\( \lim_{x \to (\frac{\pi}{2})^+} \tan x = -\infty \) Since the left-hand limit and the right-hand limit are not equal (one goes to \( +\infty \) and the other to \( -\infty \)), the limit of the function as \( x \) approaches \( \frac{\pi}{2} \) does not exist. Therefore, \( \lim_{x \to \frac{\pi}{2}} \tan x \) does not exist.In simple words: When you follow the graph of \( y = \tan x \) closer to \( x = \frac{\pi}{2} \), the line goes up to infinity on one side and down to negative infinity on the other side. Because it does not settle on one single value, the limit does not exist at this point.

🎯 Exam Tip: Remember that the tangent function has vertical asymptotes at \( x = \frac{\pi}{2} + n\pi \) where \( n \) is an integer. At these points, the limit will typically not exist because the function tends to positive or negative infinity on different sides.

 

Question 16. Sketch the graph of \( f \), then identify the values of \( x_0 \) for which \( \lim_{x \to x_0} f(x) \) exists, where \( f(x) = \begin{cases} x^2, & x \le 2 \\ 8 - 2x, & 2 < x < 4 \\ 4, & x \ge 4 \end{cases} \)

Answer:Let's evaluate the function at the critical points \( x=2 \) and \( x=4 \), as these are where the function definition changes. **At \( x=2 \):** Left-hand limit: \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4 \) Right-hand limit: \( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (8 - 2x) = 8 - 2(2) = 8 - 4 = 4 \) Since the left-hand limit equals the right-hand limit (both are 4), the limit \( \lim_{x \to 2} f(x) \) exists and is equal to 4. Also, \( f(2) = 2^2 = 4 \), so the function is continuous at \( x=2 \). **At \( x=4 \):** Left-hand limit: \( \lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} (8 - 2x) = 8 - 2(4) = 8 - 8 = 0 \) Right-hand limit: \( \lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} 4 = 4 \) Since the left-hand limit (0) is not equal to the right-hand limit (4), the limit \( \lim_{x \to 4} f(x) \) does not exist. There is a jump discontinuity at \( x=4 \). **For all other points:** * For \( x < 2 \), \( f(x) = x^2 \) is a polynomial, which is continuous everywhere. So, \( \lim_{x \to x_0} f(x) \) exists for all \( x_0 < 2 \). * For \( 2 < x < 4 \), \( f(x) = 8 - 2x \) is a linear function, which is continuous everywhere. So, \( \lim_{x \to x_0} f(x) \) exists for all \( 2 < x_0 < 4 \). * For \( x > 4 \), \( f(x) = 4 \) is a constant function, which is continuous everywhere. So, \( \lim_{x \to x_0} f(x) \) exists for all \( x_0 > 4 \). Based on this analysis, the limit \( \lim_{x \to x_0} f(x) \) exists for all real numbers \( x_0 \) except for \( x_0 = 4 \).In simple words: We check the function at the points where its rule changes. At \( x=2 \), the function smoothly connects, so the limit exists. At \( x=4 \), the function jumps, so the limit does not exist there. For all other numbers, the function is smooth, so the limit exists everywhere except at \( x=4 \).

🎯 Exam Tip: To find where a piecewise function's limit exists, always check the points where the definition changes. Calculate both the left-hand and right-hand limits at these points. If they match, the limit exists; otherwise, it does not.

 

Question 17. Verify the existence of \( \lim_{x \to 0} f(x) \), where \( f(x) = \begin{cases} \sin x, & x < 0 \\ 1 - \cos x, & 0 \le x \le \pi \\ \cos x, & x > \pi \end{cases} \)

Answer:To verify the existence of \( \lim_{x \to 0} f(x) \), we need to check the left-hand limit and the right-hand limit at \( x=0 \). **Left-hand limit at \( x=0 \):** For \( x < 0 \), \( f(x) = \sin x \). \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sin x = \sin(0) = 0 \) **Right-hand limit at \( x=0 \):** For \( 0 \le x \le \pi \), \( f(x) = 1 - \cos x \). \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1 - \cos x) = 1 - \cos(0) = 1 - 1 = 0 \) Since the left-hand limit \( (0) \) is equal to the right-hand limit \( (0) \), the limit \( \lim_{x \to 0} f(x) \) exists and is equal to 0. The value of the function at \( x=0 \) is \( f(0) = 1 - \cos(0) = 1 - 1 = 0 \). This shows the function is continuous at \( x=0 \). **Additionally, let's check the point \( x = \pi \) as well, since it is a boundary point for the function definition.** **Left-hand limit at \( x=\pi \):** For \( 0 \le x \le \pi \), \( f(x) = 1 - \cos x \). \( \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (1 - \cos x) = 1 - \cos(\pi) = 1 - (-1) = 1 + 1 = 2 \) **Right-hand limit at \( x=\pi \):** For \( x > \pi \), \( f(x) = \cos x \). \( \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} \cos x = \cos(\pi) = -1 \) Since the left-hand limit \( (2) \) is not equal to the right-hand limit \( (-1) \) at \( x=\pi \), the limit \( \lim_{x \to \pi} f(x) \) does not exist.In simple words: To see if a limit exists at a point where the function's rule changes, we check both sides. At \( x=0 \), the function smoothly connects from both sides, so the limit is 0. However, at \( x=\pi \), the function values from the left and right sides are different, so the limit does not exist there.

🎯 Exam Tip: For piecewise functions, always check the limits at the boundary points of each interval. A limit exists only if the left-hand limit and the right-hand limit are equal at that point. Also, remember standard trigonometric values like \( \sin 0 = 0 \), \( \cos 0 = 1 \), and \( \cos \pi = -1 \).

 

Question 18. Sketch the graph of a function \( f \) that satisfies the given values:
(i) \( f(0) \) is defined, \( \lim _{x \rightarrow 0} f(x) = 4 \), \( f(2) = 6 \), \( \lim _{x \rightarrow 2} f(x) = 3 \)
Answer:

In simple words: The graph shows how the function \( f(x) \) behaves. At \( x=0 \), the function value is 4, and the graph smoothly reaches 4. At \( x=2 \), the function value is 6, but the graph approaches 3 from the left side, indicating a jump.

🎯 Exam Tip: When sketching graphs for limits, always mark filled circles for defined points and open circles for limits that are not the function value.

(ii) \( f(-2) = 0 \), \( f(2) = 0 \), \( \lim _{x \rightarrow -2} f(x) = 0 \), \( \lim _{x \rightarrow 2} f(x) \) does not exist.
Answer:

In simple words: The graph shows that at \( x=-2 \), the function value is 0 and the graph passes through it. However, at \( x=2 \), the graph jumps from one level to another, meaning the limit does not exist, even though \( f(2) \) is defined as 0.

🎯 Exam Tip: Discontinuities, where the graph suddenly jumps or has a hole, are key indicators of where limits may not exist, even if the function is defined at that point.

 

Question 19. If \( \lim _{x \rightarrow 8} f(x) = 25 \), what does this imply about the function's behavior around \( x=8 \)?
Answer: If \( \lim _{x \rightarrow 8} f(x) = 25 \), it means that as \( x \) gets closer and closer to 8 from both the left side and the right side, the value of the function \( f(x) \) gets closer and closer to 25. According to the definition of a limit, this implies that the left-hand limit and the right-hand limit are equal to 25. So, \( f(8^{-}) = f(8^{+}) = 25 \).
In simple words: When the limit of a function at a point is 25, it means the function's value gets very close to 25 as you approach that point from either side.

🎯 Exam Tip: The existence of a limit at a point means that the left-hand limit and the right-hand limit are both equal to that specific value.

 

Question 20. If \( f(2) = 4 \), can you conclude anything about the limit of \( f(x) \) as \( x \) approaches 2?
Answer: No, if \( f(2) = 4 \), we cannot conclude anything definitive about \( \lim _{x \rightarrow 2} f(x) \). The value of the function at a specific point does not necessarily tell us how the function behaves as \( x \) approaches that point. The limit might exist and be equal to 4, or it might exist and be a different value, or it might not exist at all (e.g., if there's a jump discontinuity).
In simple words: Just knowing the function's value at a point doesn't tell you what its limit is there. The limit depends on what happens *near* the point, not *at* the point itself.

🎯 Exam Tip: A function can be defined at a point, but its limit at that point may be different or may not exist. Always remember that limits describe behavior *around* a point.

 

Question 21. If the limit of \( f(x) \) as \( x \) approaches 2 is 4, can you conclude anything about \( f(2) \)? Explain reasoning.
Answer: If \( \lim _{x \rightarrow 2} f(x) = 4 \), this means that as \( x \) approaches 2 from both the left and the right (i.e., \( \lim _{x \rightarrow 2^{-}} f(x) = 4 \) and \( \lim _{x \rightarrow 2^{+}} f(x) = 4 \)), the function values approach 4. However, we cannot conclude anything about the actual value of \( f(2) \) itself. The function \( f(x) \) could be defined as 4 at \( x=2 \), or it could be defined as a different value, or it might not be defined at \( x=2 \) at all (a hole in the graph). The limit only describes the approach, not the value at the point.
In simple words: Knowing that a function's limit is 4 as \( x \) gets close to 2 does not tell us what the function's exact value is *at* \( x=2 \). It could be 4, something else, or undefined.

🎯 Exam Tip: The existence and value of a limit are independent of the function's value at the point in question. A function can have a limit even if it's undefined at that point, or if its value is different from the limit.

 

Question 22. Evaluate: \( \lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3} \) if it exists by finding the left and right-hand limits at \( x=3 \).
Answer: To evaluate the limit, we first simplify the expression: \[ \lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3} \] We know that \( x^2 - 9 = (x - 3)(x + 3) \). So, substitute this into the expression: \[ = \lim_{x \rightarrow 3} \frac{(x - 3)(x + 3)}{x - 3} \] For \( x \neq 3 \), we can cancel out the \( (x - 3) \) terms. \[ = \lim_{x \rightarrow 3} (x + 3) \] Now, substitute \( x = 3 \) into the simplified expression to find the limit: \[ = 3 + 3 = 6 \] We can see that as \( x \) approaches 3 from values less than 3 (left-hand limit, \( x \rightarrow 3^- \)), \( x+3 \) approaches \( 3+3=6 \). Similarly, as \( x \) approaches 3 from values greater than 3 (right-hand limit, \( x \rightarrow 3^+ \)), \( x+3 \) approaches \( 3+3=6 \). Since the left-hand limit equals the right-hand limit, the limit exists and is 6. Thus, \( \lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3} = 6 \).
In simple words: We can simplify the fraction by factoring the top part. After cancelling terms, we are left with \( x+3 \). When we put 3 into this simple expression, we get 6. This means as \( x \) gets close to 3, the value of the fraction gets close to 6 from both sides.

🎯 Exam Tip: Always try to simplify rational expressions by factoring before directly substituting the limit value, especially when direct substitution results in an an indeterminate form like \( \frac{0}{0} \).

 

Question 23. Verify the existence of \( \lim _{x \rightarrow 1} f(x) \), where \[ f(x) = \begin{cases} \frac{|x - 1|}{x - 1} & \text{for } x \neq 1 \\ 0 & \text{for } x = 1 \end{cases} \]
Answer: To verify the existence of the limit as \( x \rightarrow 1 \), we need to check the left-hand limit and the right-hand limit. First, let's simplify \( \frac{|x - 1|}{x - 1} \). If \( x > 1 \), then \( x - 1 > 0 \), so \( |x - 1| = x - 1 \). Therefore, \( \frac{|x - 1|}{x - 1} = \frac{x - 1}{x - 1} = 1 \) for \( x > 1 \). If \( x < 1 \), then \( x - 1 < 0 \), so \( |x - 1| = -(x - 1) \). Therefore, \( \frac{|x - 1|}{x - 1} = \frac{-(x - 1)}{x - 1} = -1 \) for \( x < 1 \). Now, let's find the left-hand limit: \( \lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} \frac{-(x - 1)}{x - 1} \)
\( \implies \lim _{x \rightarrow 1^{-}} (-1) = -1 \) Next, let's find the right-hand limit: \( \lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} \frac{x - 1}{x - 1} \)
\( \implies \lim _{x \rightarrow 1^{+}} (1) = 1 \) Since the left-hand limit ( -1 ) is not equal to the right-hand limit ( 1 ), the limit \( \lim _{x \rightarrow 1} f(x) \) does not exist. The function has a jump discontinuity at \( x=1 \).
In simple words: This function behaves differently on each side of \( x=1 \). When \( x \) is a little less than 1, the function value is -1. When \( x \) is a little more than 1, the function value is 1. Since these two values are not the same, the limit at \( x=1 \) does not exist.

🎯 Exam Tip: For piecewise functions involving absolute values, always simplify the absolute value expression based on whether the term inside is positive or negative, then evaluate the left and right-hand limits separately.