OP Malhotra Class 11 Maths Solutions Chapter 18 Limits Exercise 18 (A)

Question 1. Show that \( \text{Lim}_{x \rightarrow 2} \frac{|x-2|}{x-2} \) does not exist.
Answer: To show that the limit does not exist, we need to check if the Left Hand Limit (L.H.L) and Right Hand Limit (R.H.L) are equal.
First, let's find the L.H.L:
\( \text{L.H.L} = \text{Lt}_{x \rightarrow 2^-} f(x) = \text{Lt}_{x \rightarrow 2^-} \frac{|x-2|}{x-2} \)
When \( x \rightarrow 2^- \), it means \( x < 2 \), so \( x-2 < 0 \).
Therefore, \( |x-2| = -(x-2) \).
\( \text{L.H.L} = \text{Lt}_{x \rightarrow 2^-} \frac{-(x-2)}{x-2} = \text{Lt}_{x \rightarrow 2^-} (-1) = -1 \)
Next, let's find the R.H.L:
\( \text{R.H.L} = \text{Lt}_{x \rightarrow 2^+} f(x) = \text{Lt}_{x \rightarrow 2^+} \frac{|x-2|}{x-2} \)
When \( x \rightarrow 2^+ \), it means \( x > 2 \), so \( x-2 > 0 \).
Therefore, \( |x-2| = (x-2) \).
\( \text{R.H.L} = \text{Lt}_{x \rightarrow 2^+} \frac{(x-2)}{x-2} = \text{Lt}_{x \rightarrow 2^+} (1) = 1 \)
Since L.H.L \( \neq \) R.H.L (because \( -1 \neq 1 \)), the limit of the function does not exist at \( x = 2 \). This indicates a sharp change in value at that point.

Alternatively, using substitution with \( h \):
For L.H.L, let \( x = 2-h \). As \( x \rightarrow 2^- \), \( h \rightarrow 0^+ \).
\( \text{L.H.L} = \text{Lt}_{h \rightarrow 0^+} \frac{|(2-h)-2|}{(2-h)-2} = \text{Lt}_{h \rightarrow 0^+} \frac{|-h|}{-h} \)
Since \( h \rightarrow 0^+ \), \( h \) is a small positive number, so \( |-h| = h \).
\( \text{L.H.L} = \text{Lt}_{h \rightarrow 0^+} \frac{h}{-h} = \text{Lt}_{h \rightarrow 0^+} (-1) = -1 \)
For R.H.L, let \( x = 2+h \). As \( x \rightarrow 2^+ \), \( h \rightarrow 0^+ \).
\( \text{R.H.L} = \text{Lt}_{h \rightarrow 0^+} \frac{|(2+h)-2|}{(2+h)-2} = \text{Lt}_{h \rightarrow 0^+} \frac{|h|}{h} \)
Since \( h \rightarrow 0^+ \), \( |h| = h \).
\( \text{R.H.L} = \text{Lt}_{h \rightarrow 0^+} \frac{h}{h} = \text{Lt}_{h \rightarrow 0^+} (1) = 1 \)
Again, L.H.L \( \neq \) R.H.L.
In simple words: When we check the value the function approaches from the left side of 2, it gives -1. But when we check from the right side of 2, it gives 1. Since these two values are not the same, the function does not have a single, clear limit at 2.

🎯 Exam Tip: For functions involving absolute values, always split into L.H.L and R.H.L and evaluate the absolute value expression based on whether \( x \) is less than or greater than the critical point.

Question 2. Show that \( \text{Lim}_{x \rightarrow \frac{\pi}{2}} \tan x \) does not exist.
Answer: To prove that the limit does not exist, we need to examine the Left Hand Limit (L.H.L) and Right Hand Limit (R.H.L).
For the L.H.L, we consider \( x \rightarrow \frac{\pi}{2}^- \). Let \( x = \frac{\pi}{2} - h \), where \( h \rightarrow 0^+ \).
\( \text{L.H.L} = \text{Lt}_{x \rightarrow \frac{\pi}{2}^-} \tan x = \text{Lt}_{h \rightarrow 0^+} \tan \left( \frac{\pi}{2} - h \right) \)
We know that \( \tan \left( \frac{\pi}{2} - h \right) = \cot h \).
\( \text{L.H.L} = \text{Lt}_{h \rightarrow 0^+} \cot h \)
As \( h \rightarrow 0^+ \), \( \cot h \rightarrow \infty \). So, L.H.L = \( \infty \).
For the R.H.L, we consider \( x \rightarrow \frac{\pi}{2}^+ \). Let \( x = \frac{\pi}{2} + h \), where \( h \rightarrow 0^+ \).
\( \text{R.H.L} = \text{Lt}_{x \rightarrow \frac{\pi}{2}^+} \tan x = \text{Lt}_{h \rightarrow 0^+} \tan \left( \frac{\pi}{2} + h \right) \)
We know that \( \tan \left( \frac{\pi}{2} + h \right) = -\cot h \).
\( \text{R.H.L} = \text{Lt}_{h \rightarrow 0^+} (-\cot h) \)
As \( h \rightarrow 0^+ \), \( -\cot h \rightarrow -\infty \). So, R.H.L = \( -\infty \).
Since the L.H.L (\( \infty \)) is not equal to the R.H.L (\( -\infty \)), the limit of \( \tan x \) as \( x \rightarrow \frac{\pi}{2} \) does not exist. This is a common behavior for vertical asymptotes in trigonometric functions.
In simple words: When you get very close to \( \frac{\pi}{2} \) from the left side, the tangent function goes up to a very big positive number (infinity). When you get very close from the right side, it goes to a very big negative number (negative infinity). Because it doesn't settle on one number, the limit does not exist.

🎯 Exam Tip: Remember the behavior of trigonometric functions near their asymptotes. For tangent, \( \frac{\pi}{2} \) is a critical point where the function values go towards positive or negative infinity.

Question 3. If \( f(x) = \begin{cases} x^2 + 4 & \text{for } x < 2 \\ x^3 & \text{for } x > 2 \end{cases} \), find \( \text{Lim}_{x \rightarrow 2} f(x) \).
Answer: To find the limit of the function \( f(x) \) as \( x \rightarrow 2 \), we need to check if the Left Hand Limit (L.H.L) and Right Hand Limit (R.H.L) are equal.
First, let's find the L.H.L:
\( \text{L.H.L} = \text{Lt}_{x \rightarrow 2^-} f(x) \)
When \( x < 2 \), the function definition is \( f(x) = x^2 + 4 \).
So, \( \text{L.H.L} = \text{Lt}_{x \rightarrow 2^-} (x^2 + 4) \)
We can substitute \( x = 2 \) directly into the expression for a polynomial function:
\( \text{L.H.L} = (2)^2 + 4 = 4 + 4 = 8 \)
Next, let's find the R.H.L:
\( \text{R.H.L} = \text{Lt}_{x \rightarrow 2^+} f(x) \)
When \( x > 2 \), the function definition is \( f(x) = x^3 \).
So, \( \text{R.H.L} = \text{Lt}_{x \rightarrow 2^+} (x^3) \)
Substitute \( x = 2 \) directly:
\( \text{R.H.L} = (2)^3 = 8 \)
Since L.H.L = R.H.L (both are 8), the limit of \( f(x) \) as \( x \rightarrow 2 \) exists and is equal to 8. This shows the function approaches the same value from both sides, even if it's defined differently.
In simple words: When we look at numbers slightly smaller than 2, the function acts like \( x^2+4 \), and its value gets close to 8. When we look at numbers slightly bigger than 2, it acts like \( x^3 \), and its value also gets close to 8. Since both sides point to 8, the limit at \( x=2 \) is 8.

🎯 Exam Tip: For piecewise functions, always use the correct function definition for the L.H.L (for values less than the point) and the R.H.L (for values greater than the point). If both limits are equal, the overall limit exists.

Question 4. If \( f(x) = \begin{cases} \frac{x}{|x|} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases} \), find \( \text{Lim}_{x \rightarrow 0} f(x) \).
Answer: To determine if the limit of \( f(x) \) as \( x \rightarrow 0 \) exists, we must evaluate the Left Hand Limit (L.H.L) and Right Hand Limit (R.H.L).
First, let's find the L.H.L:
\( \text{L.H.L} = \text{Lt}_{x \rightarrow 0^-} f(x) = \text{Lt}_{x \rightarrow 0^-} \frac{x}{|x|} \)
When \( x \rightarrow 0^- \), it means \( x < 0 \). For negative numbers, \( |x| = -x \).
So, \( \text{L.H.L} = \text{Lt}_{x \rightarrow 0^-} \frac{x}{-x} = \text{Lt}_{x \rightarrow 0^-} (-1) = -1 \)
Next, let's find the R.H.L:
\( \text{R.H.L} = \text{Lt}_{x \rightarrow 0^+} f(x) = \text{Lt}_{x \rightarrow 0^+} \frac{x}{|x|} \)
When \( x \rightarrow 0^+ \), it means \( x > 0 \). For positive numbers, \( |x| = x \).
So, \( \text{R.H.L} = \text{Lt}_{x \rightarrow 0^+} \frac{x}{x} = \text{Lt}_{x \rightarrow 0^+} (1) = 1 \)
Since the L.H.L (\( -1 \)) is not equal to the R.H.L (\( 1 \)), the limit of \( f(x) \) as \( x \rightarrow 0 \) does not exist. This function is often called the signum function, which clearly shows a jump discontinuity at zero.
In simple words: When we get close to 0 from the left (negative numbers), the function's value is -1. When we get close to 0 from the right (positive numbers), the function's value is 1. Since these two values are different, there is no single limit at 0.

🎯 Exam Tip: Functions involving \( \frac{x}{|x|} \) typically have different left and right limits at \( x=0 \) due to the definition of the absolute value for positive and negative numbers. This is a common way to demonstrate a limit that does not exist.

Question 5. Does \( \text{Lim}_{x \rightarrow 0} f(x) \) exist if \( f(x) = \begin{cases} x & \text{when } x < 0 \\ 0 & \text{when } x = 0 \\ x^2 & \text{when } x > 0 \end{cases} \)?
Answer: To check if the limit of \( f(x) \) exists as \( x \rightarrow 0 \), we need to compare the Left Hand Limit (L.H.L) and Right Hand Limit (R.H.L).
First, let's find the L.H.L:
\( \text{L.H.L} = \text{Lt}_{x \rightarrow 0^-} f(x) \)
When \( x < 0 \), the function definition is \( f(x) = x \).
So, \( \text{L.H.L} = \text{Lt}_{x \rightarrow 0^-} (x) \)
Substituting \( x = 0 \):
\( \text{L.H.L} = 0 \)
Next, let's find the R.H.L:
\( \text{R.H.L} = \text{Lt}_{x \rightarrow 0^+} f(x) \)
When \( x > 0 \), the function definition is \( f(x) = x^2 \).
So, \( \text{R.H.L} = \text{Lt}_{x \rightarrow 0^+} (x^2) \)
Substituting \( x = 0 \):
\( \text{R.H.L} = (0)^2 = 0 \)
Since L.H.L = R.H.L (both are 0), the limit of \( f(x) \) as \( x \rightarrow 0 \) exists and is equal to 0. Although the function has a specific value \( f(0)=0 \) at \( x=0 \), the limit only cares about the values the function *approaches*.
In simple words: As we get closer to 0 from the left side, the function's value gets closer to 0. As we get closer to 0 from the right side, the function's value also gets closer to 0. Since both sides meet at 0, the limit exists and is 0.

🎯 Exam Tip: When evaluating limits for piecewise functions, remember that the value of the function *at* the point (e.g., \( f(0) \) here) is separate from the limit. The limit only concerns values *near* the point.

Question 6. Show that \( \text{Lim}_{x \rightarrow 0} \frac{1}{x} \) does not exist.
Answer: To show that the limit of \( \frac{1}{x} \) as \( x \rightarrow 0 \) does not exist, we must examine its Left Hand Limit (L.H.L) and Right Hand Limit (R.H.L).
First, let's find the L.H.L:
\( \text{L.H.L} = \text{Lt}_{x \rightarrow 0^-} \frac{1}{x} \)
When \( x \rightarrow 0^- \), \( x \) is a very small negative number (e.g., -0.001).
So, \( \frac{1}{x} \) becomes a very large negative number.
Therefore, \( \text{L.H.L} = -\infty \).
Next, let's find the R.H.L:
\( \text{R.H.L} = \text{Lt}_{x \rightarrow 0^+} \frac{1}{x} \)
When \( x \rightarrow 0^+ \), \( x \) is a very small positive number (e.g., 0.001).
So, \( \frac{1}{x} \) becomes a very large positive number.
Therefore, \( \text{R.H.L} = \infty \).
Since the L.H.L (\( -\infty \)) is not equal to the R.H.L (\( \infty \)), the limit of \( \frac{1}{x} \) as \( x \rightarrow 0 \) does not exist. This is a classic example of a vertical asymptote where the function "blows up" in opposite directions.
In simple words: If you divide 1 by a number very close to zero but negative, the answer is a huge negative number. If you divide 1 by a number very close to zero but positive, the answer is a huge positive number. Since the values go to different infinities, there is no single limit.

🎯 Exam Tip: Always analyze the sign of the denominator when dealing with limits that result in division by zero. If the expression approaches \( \frac{1}{0} \), it's likely that the limit will be \( \infty \), \( -\infty \), or D.N.E. depending on the signs from left and right.

Question 7. Show that \( \text{Lim}_{x \rightarrow 0} e^{-\frac{1}{x}} \) does not exist.
Answer: To prove that the limit of \( e^{-\frac{1}{x}} \) as \( x \rightarrow 0 \) does not exist, we need to evaluate the Left Hand Limit (L.H.L) and Right Hand Limit (R.H.L).
First, let's find the L.H.L:
\( \text{L.H.L} = \text{Lt}_{x \rightarrow 0^-} e^{-\frac{1}{x}} \)
Let \( x = 0 - h \), where \( h \rightarrow 0^+ \).
Then, \( -\frac{1}{x} = -\frac{1}{(-h)} = \frac{1}{h} \).
As \( h \rightarrow 0^+ \), \( \frac{1}{h} \rightarrow \infty \).
So, \( \text{L.H.L} = \text{Lt}_{h \rightarrow 0^+} e^{\frac{1}{h}} = e^{\infty} = \infty \).
Next, let's find the R.H.L:
\( \text{R.H.L} = \text{Lt}_{x \rightarrow 0^+} e^{-\frac{1}{x}} \)
Let \( x = 0 + h \), where \( h \rightarrow 0^+ \).
Then, \( -\frac{1}{x} = -\frac{1}{h} \).
As \( h \rightarrow 0^+ \), \( \frac{1}{h} \rightarrow \infty \), so \( -\frac{1}{h} \rightarrow -\infty \).
So, \( \text{R.H.L} = \text{Lt}_{h \rightarrow 0^+} e^{-\frac{1}{h}} = e^{-\infty} = 0 \).
Since the L.H.L (\( \infty \)) is not equal to the R.H.L (\( 0 \)), the limit of \( e^{-\frac{1}{x}} \) as \( x \rightarrow 0 \) does not exist. This function has a fascinating behavior where it approaches zero from one side and grows infinitely from the other.
In simple words: When \( x \) is a tiny negative number, \( -1/x \) becomes a very large positive number, so \( e^{-1/x} \) becomes extremely big (infinity). When \( x \) is a tiny positive number, \( -1/x \) becomes a very large negative number, so \( e^{-1/x} \) becomes almost zero. Because the function goes to different values from each side, the limit does not exist.

🎯 Exam Tip: When evaluating limits involving exponential functions like \( e^{\frac{1}{x}} \) or \( e^{-\frac{1}{x}} \) near \( x=0 \), remember that \( e^{\infty} \rightarrow \infty \) and \( e^{-\infty} \rightarrow 0 \). The sign of the exponent will be crucial for determining the limit.

Question 8. Show that \( \text{Lim}_{x \rightarrow 0} \sin \frac{1}{x} \) does not exist.
Answer: To show that the limit of \( \sin \frac{1}{x} \) as \( x \rightarrow 0 \) does not exist, we need to understand the behavior of the function as \( x \) approaches 0.
As \( x \rightarrow 0 \), the argument \( \frac{1}{x} \) either goes to \( \infty \) (from \( x \rightarrow 0^+ \)) or \( -\infty \) (from \( x \rightarrow 0^- \)).
The sine function, \( \sin \theta \), oscillates between -1 and 1 for any real value of \( \theta \).
Consider the L.H.L:
\( \text{L.H.L} = \text{Lt}_{x \rightarrow 0^-} \sin \frac{1}{x} \)
As \( x \rightarrow 0^- \), \( \frac{1}{x} \) approaches \( -\infty \). The sine function will continue to oscillate between -1 and 1 infinitely many times as its input goes towards \( -\infty \). It does not settle on a single value.
Consider the R.H.L:
\( \text{R.H.L} = \text{Lt}_{x \rightarrow 0^+} \sin \frac{1}{x} \)
As \( x \rightarrow 0^+ \), \( \frac{1}{x} \) approaches \( \infty \). Similarly, the sine function will continue to oscillate between -1 and 1 infinitely many times as its input goes towards \( \infty \). It does not approach a unique value.
Since the function keeps oscillating and does not approach a single, definite value from either the left or the right side of 0, the limit of \( \sin \frac{1}{x} \) as \( x \rightarrow 0 \) does not exist. This oscillating behavior is characteristic of such limits.
In simple words: As \( x \) gets closer and closer to 0, the value \( 1/x \) gets infinitely large or small. The sine function then rapidly swings back and forth between -1 and 1, never settling on just one number. Because it keeps jumping around, there is no single limit.

🎯 Exam Tip: For oscillating functions like \( \sin(\frac{1}{x}) \) or \( \cos(\frac{1}{x}) \) as \( x \rightarrow 0 \), the key is to recognize that the argument of the trigonometric function approaches infinity, causing infinite oscillations. This prevents the limit from existing.

Question 9. If \( f \) is an even function, then prove that \( \text{Lim}_{x \rightarrow 0^-} f(x) = \text{Lim}_{x \rightarrow 0^+} f(x) \), whenever they exist.
Answer: Given that \( f \) is an even function, its definition is \( f(-x) = f(x) \) for all \( x \) in its domain. We need to prove that if the limits exist, the Left Hand Limit (L.H.L) and Right Hand Limit (R.H.L) at \( x=0 \) are equal.
First, let's consider the L.H.L:
\( \text{L.H.L} = \text{Lt}_{x \rightarrow 0^-} f(x) \)
Let \( x = -h \), where \( h \rightarrow 0^+ \). (As \( x \) approaches 0 from the left, \( x \) is negative, so \( h \) is positive and approaches 0.)
Substituting \( x = -h \) into the L.H.L expression:
\( \text{L.H.L} = \text{Lt}_{h \rightarrow 0^+} f(-h) \)
Since \( f \) is an even function, we know that \( f(-h) = f(h) \).
So, \( \text{L.H.L} = \text{Lt}_{h \rightarrow 0^+} f(h) \)
By definition, \( \text{Lt}_{h \rightarrow 0^+} f(h) \) is the Right Hand Limit of \( f(x) \) as \( x \rightarrow 0 \) (since \( h \) acts like a positive \( x \) approaching 0).
Thus, \( \text{L.H.L} = \text{Lt}_{x \rightarrow 0^+} f(x) = \text{R.H.L} \).
Therefore, if \( f \) is an even function and the limits exist, then \( \text{Lim}_{x \rightarrow 0^-} f(x) = \text{Lim}_{x \rightarrow 0^+} f(x) \). This property ensures that the function behaves symmetrically around the y-axis, and if a limit exists at zero, it must be approached identically from both sides.
In simple words: An even function looks the same on both sides of the y-axis, like a mirror image. Because of this, when you look at what the function gets close to as you come from the left side of 0, it will be the exact same as what it gets close to when you come from the right side of 0.

🎯 Exam Tip: The property of even functions \( f(-x) = f(x) \) is fundamental. Use variable substitution (e.g., \( x = -h \)) to transform the L.H.L into an R.H.L in terms of \( h \), then apply the even function property to complete the proof.

Question 10. For what values of \( p \) does the \( \text{Lim}_{x \rightarrow 1} f(x) \) exist, where \( f(x) \) is defined by the rule \( f(x) = \begin{cases} 2px + 3 & \text{if } x < 1 \\ 1 - px^2 & \text{if } x > 1 \end{cases} \)?
Answer: For the limit of \( f(x) \) to exist as \( x \rightarrow 1 \), the Left Hand Limit (L.H.L) must be equal to the Right Hand Limit (R.H.L).
First, let's find the L.H.L:
\( \text{L.H.L} = \text{Lt}_{x \rightarrow 1^-} f(x) \)
When \( x < 1 \), the function definition is \( f(x) = 2px + 3 \).
So, \( \text{L.H.L} = \text{Lt}_{x \rightarrow 1^-} (2px + 3) \)
Substitute \( x = 1 \):
\( \text{L.H.L} = 2p(1) + 3 = 2p + 3 \)
Next, let's find the R.H.L:
\( \text{R.H.L} = \text{Lt}_{x \rightarrow 1^+} f(x) \)
When \( x > 1 \), the function definition is \( f(x) = 1 - px^2 \).
So, \( \text{R.H.L} = \text{Lt}_{x \rightarrow 1^+} (1 - px^2) \)
Substitute \( x = 1 \):
\( \text{R.H.L} = 1 - p(1)^2 = 1 - p \)
For the limit to exist, L.H.L = R.H.L.
\( 2p + 3 = 1 - p \)
Now, we solve for \( p \):
\( 2p + p = 1 - 3 \)
\( 3p = -2 \)
\( p = -\frac{2}{3} \)
Therefore, the limit of \( f(x) \) as \( x \rightarrow 1 \) exists only when \( p = -\frac{2}{3} \). This value makes the two pieces of the function meet at the point \( x=1 \).
In simple words: For the function to have a smooth meeting point at \( x=1 \), the part of the function for numbers smaller than 1 must meet up with the part for numbers bigger than 1. By setting these two parts equal when \( x=1 \), we can find the exact value of \( p \) that makes them match.

🎯 Exam Tip: When finding a constant (like \( p \)) for which a limit exists for a piecewise function, always set the L.H.L equal to the R.H.L at the point where the definition changes. Then, solve the resulting equation for the constant.