RBSE Solutions Class 11 Maths Chapter 10 Limits and Derivatives Exercise 10.1

Rajasthan Board RBSE Class 11 Maths Chapter 10 Limits and Derivatives Ex 10.1

Question 1. Show that left and right limits of function \( f(x) = \frac { \log_{e} x }{ x-1 } \) at x = 1 are equal and their value is 1.
Answer:
For the right-hand limit at \( x = 1 \), we use \( x = 1 + h \) where \( h \to 0 \).
\[ \text{R.H.L.} = \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1+h) \]
\[ = \lim_{h \to 0} \frac { \log_{e} (1+h) }{ (1+h) - 1 } = \lim_{h \to 0} \frac { \log_{e} (1+h) }{ h } \] Using the series expansion for \( \log_{e} (1+h) = h - \frac { h^2 }{ 2 } + \frac { h^3 }{ 3 } - ... \)
\[ = \lim_{h \to 0} \frac { h - \frac { h^2 }{ 2 } + \frac { h^3 }{ 3 } - ... }{ h } \]
\[ = \lim_{h \to 0} \left[ 1 - \frac { h }{ 2 } + \frac { h^2 }{ 3 } - ... \right] \]
\[ = 1 - 0 + 0 - ... = 1 \quad ...(i) \] For the left-hand limit at \( x = 1 \), we use \( x = 1 - h \) where \( h \to 0 \).
\[ \text{L.H.L.} = \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1-h) \]
\[ = \lim_{h \to 0} \frac { \log_{e} (1-h) }{ (1-h) - 1 } = \lim_{h \to 0} \frac { \log_{e} (1-h) }{ -h } \] Using the series expansion for \( \log_{e} (1-h) = -h - \frac { h^2 }{ 2 } - \frac { h^3 }{ 3 } - ... \)
\[ = \lim_{h \to 0} \frac { -h - \frac { h^2 }{ 2 } - \frac { h^3 }{ 3 } - ... }{ -h } \]
\[ = \lim_{h \to 0} \left[ 1 + \frac { h }{ 2 } + \frac { h^2 }{ 3 } + ... \right] \]
\[ = 1 + 0 + 0 + ... = 1 \quad ...(ii) \] From equations (i) and (ii), the left-hand limit is equal to the right-hand limit.
Therefore, L.H.L. = R.H.L. = 1.
This means the limit of the function at \( x=1 \) exists and is equal to 1.
In simple words: We check the function's value as we get closer to 1 from both sides. If both approaches give the same number, then the limit exists and equals that number. Here, both sides approach 1, so the limit is 1.

🎯 Exam Tip: Remember to use the appropriate series expansions for \( \log_{e}(1+h) \) and \( \log_{e}(1-h) \) when evaluating limits of logarithmic functions.

Question 2. Examine the existence of the limit of the function \( f(x) = \frac { |x| }{ x } \) at \( x=0 \).
Answer:
We need to find the right-hand limit (R.H.L.) and the left-hand limit (L.H.L.) at \( x = 0 \).
For the right-hand limit at \( x = 0 \), we use \( x = 0 + h \) where \( h > 0 \).
\[ \text{R.H.L.} = \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0+h) \]
\[ = \lim_{h \to 0} \frac { |0+h| }{ 0+h } = \lim_{h \to 0} \frac { |h| }{ h } \] Since \( h > 0 \), \( |h| = h \).
\[ = \lim_{h \to 0} \frac { h }{ h } = \lim_{h \to 0} 1 = 1 \quad ...(i) \] For the left-hand limit at \( x = 0 \), we use \( x = 0 - h \) where \( h > 0 \). (Note: sometimes \( h < 0 \) notation is used, but for simplicity, we keep \( h \to 0^+ \) and adjust the function input).
\[ \text{L.H.L.} = \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0-h) \]
\[ = \lim_{h \to 0} \frac { |0-h| }{ 0-h } = \lim_{h \to 0} \frac { |-h| }{ -h } \] Since \( h > 0 \), \( |-h| = h \).
\[ = \lim_{h \to 0} \frac { h }{ -h } = \lim_{h \to 0} (-1) = -1 \quad ...(ii) \] From equations (i) and (ii), we see that L.H.L. \( \neq \) R.H.L. (since \( 1 \neq -1 \)).
Therefore, the limit of the function \( f(x) = \frac { |x| }{ x } \) does not exist at \( x = 0 \).
In simple words: When we get close to zero from the positive side, the function gives 1. When we get close to zero from the negative side, it gives -1. Since these two values are different, the function does not have a single, clear limit at zero.

🎯 Exam Tip: For functions involving absolute values, always consider both the positive and negative cases (right-hand and left-hand limits) because the absolute value definition changes at the point of interest.

Question 3. Prove that at x = 0, limits of function \( f(x) = | x | + | x - 1| \) exists.
Answer:
We need to find the right-hand limit (R.H.L.) and the left-hand limit (L.H.L.) at \( x = 0 \).
For the right-hand limit at \( x = 0 \), we use \( x = 0 + h \) where \( h > 0 \).
\[ \text{R.H.L.} = \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0+h) \]
\[ = \lim_{h \to 0} [|0+h| + |(0+h)-1|] \]
\[ = \lim_{h \to 0} [|h| + |h-1|] \] Since \( h > 0 \) and is very small, \( |h| = h \) and \( |h-1| = -(h-1) = 1-h \).
\[ = \lim_{h \to 0} [h + (1-h)] = \lim_{h \to 0} [1] = 1 \quad ...(i) \] For the left-hand limit at \( x = 0 \), we use \( x = 0 - h \) where \( h > 0 \).
\[ \text{L.H.L.} = \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0-h) \]
\[ = \lim_{h \to 0} [|0-h| + |(0-h)-1|] \]
\[ = \lim_{h \to 0} [|-h| + |-h-1|] \] Since \( h > 0 \) and is very small, \( |-h| = h \) and \( |-h-1| = |-(h+1)| = h+1 \).
\[ = \lim_{h \to 0} [h + (h+1)] = \lim_{h \to 0} [2h+1] = 2(0)+1 = 1 \quad ...(ii) \] From equations (i) and (ii), L.H.L. = R.H.L. = 1.
Thus, the limit of the function \( f(x) = |x| + |x-1| \) exists at \( x = 0 \) and its value is 1.
In simple words: We tested the function as we got closer to 0 from both sides. Both approaches resulted in the number 1. Because the results match, the limit exists at \( x = 0 \) and is 1.

🎯 Exam Tip: When evaluating limits of functions with multiple absolute value terms, carefully determine the sign of each expression inside the absolute value for the specific limit direction (left or right).

Question 4. Prove that at x = 2, limits of function does not exists.
\[ f(x) = \begin{cases} x^2+x+1, & \text{if } x \geq 2 \\ x, & \text{if } x < 2 \end{cases} \]
Answer:
We need to find the right-hand limit (R.H.L.) and the left-hand limit (L.H.L.) at \( x = 2 \).
For the right-hand limit at \( x = 2 \), we use \( x = 2 + h \) where \( h > 0 \). Since \( x \geq 2 \), we use the first part of the function definition: \( f(x) = x^2+x+1 \).
\[ \text{R.H.L.} = \lim_{x \to 2^+} f(x) = \lim_{h \to 0} f(2+h) \]
\[ = \lim_{h \to 0} [(2+h)^2 + (2+h) + 1] \]
\[ = \lim_{h \to 0} [4 + 4h + h^2 + 2 + h + 1] \]
\[ = \lim_{h \to 0} [h^2 + 5h + 7] \] Now, substitute \( h = 0 \):
\[ = 0^2 + 5(0) + 7 = 7 \quad ...(i) \] For the left-hand limit at \( x = 2 \), we use \( x = 2 - h \) where \( h > 0 \). Since \( x < 2 \), we use the second part of the function definition: \( f(x) = x \).
\[ \text{L.H.L.} = \lim_{x \to 2^-} f(x) = \lim_{h \to 0} f(2-h) \]
\[ = \lim_{h \to 0} [2-h] \] Now, substitute \( h = 0 \):
\[ = 2 - 0 = 2 \quad ...(ii) \] From equations (i) and (ii), we see that L.H.L. \( \neq \) R.H.L. (since \( 2 \neq 7 \)).
Therefore, the limit of the function \( f(x) \) does not exist at \( x = 2 \).
In simple words: When we approach the number 2 from the right side, the function's value gets close to 7. But when we approach 2 from the left side, the function's value gets close to 2. Since these two values are different, the limit does not exist at \( x = 2 \).

🎯 Exam Tip: For piecewise functions, carefully identify which part of the function definition applies to the right-hand limit and which applies to the left-hand limit at the point of interest.

Question 5. Examine the existence of the limit of the function \( f(x) = x \cos \left( \frac { 1 }{ x } \right) \) at \( x=0 \).
Answer:
We need to find the right-hand limit (R.H.L.) and the left-hand limit (L.H.L.) at \( x = 0 \).
For the right-hand limit at \( x = 0 \), we use \( x = 0 + h \) where \( h > 0 \).
\[ \text{R.H.L.} = \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0+h) \]
\[ = \lim_{h \to 0} [(0+h) \cos \left( \frac { 1 }{ 0+h } \right) ] \]
\[ = \lim_{h \to 0} [h \cos \left( \frac { 1 }{ h } \right) ] \] As \( h \to 0 \), \( \cos \left( \frac { 1 }{ h } \right) \) oscillates between -1 and 1. So, it is a finite rational number between -1 and 1.
\[ = 0 \times (\text{any rational number between -1 and 1}) \]
\[ = 0 \quad ...(i) \] For the left-hand limit at \( x = 0 \), we use \( x = 0 - h \) where \( h > 0 \).
\[ \text{L.H.L.} = \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0-h) \]
\[ = \lim_{h \to 0} [(0-h) \cos \left( \frac { 1 }{ 0-h } \right) ] \]
\[ = \lim_{h \to 0} [-h \cos \left( -\frac { 1 }{ h } \right) ] \] Since \( \cos(-\theta) = \cos(\theta) \), this becomes:
\[ = \lim_{h \to 0} [-h \cos \left( \frac { 1 }{ h } \right) ] \] As \( h \to 0 \), \( \cos \left( \frac { 1 }{ h } \right) \) oscillates between -1 and 1. So, it is a finite rational number between -1 and 1.
\[ = -0 \times (\text{any rational number between -1 and 1}) \]
\[ = 0 \quad ...(ii) \] From equations (i) and (ii), L.H.L. = R.H.L. = 0.
Therefore, the limit of the function \( f(x) = x \cos \left( \frac { 1 }{ x } \right) \) exists at \( x = 0 \) and its value is 0.
In simple words: Even though \( \cos(1/x) \) jumps around a lot near zero, when you multiply it by \( x \) (which goes to zero), the whole expression gets squeezed to zero. This happens from both the left and right sides, so the limit exists and is 0.

🎯 Exam Tip: For limits involving \( x \sin(1/x) \) or \( x \cos(1/x) \) as \( x \to 0 \), use the Squeeze Theorem. Since \( -1 \leq \sin(1/x) \leq 1 \) and \( -1 \leq \cos(1/x) \leq 1 \), multiplying by \( x \) (for \( x > 0 \)) gives \( -x \leq x \sin(1/x) \leq x \). As \( x \to 0 \), \( -x \to 0 \) and \( x \to 0 \), so the limit is 0.