Class 11 Mathematics Chapter 12 Limits and Derivatives All Exercise Solutions
Complete step-by-step solutions for NCERT Class 11 Mathematics Chapter 12 - Limits and Derivatives.
Ex 12.1 - Question 1
Question: $\lim_{x \to 3} x + 3$
Solution:
To evaluate the limit:
$$\lim_{x \to 3} (x + 3)$$
Direct substitution of $x = 3$ gives:
$$= 3 + 3 = 6$$
Answer: $6$
Ex 12.1 - Question 2
Question: $\lim_{x \to \pi} \left(x - \frac{22}{7}\right)$
Solution:
To evaluate the limit:
$$\lim_{x \to \pi} \left(x - \frac{22}{7}\right)$$
Direct substitution of $x = \pi$ gives:
$$= \pi - \frac{22}{7}$$
Answer: $\left(\pi - \frac{22}{7}\right)$
Ex 12.1 - Question 3
Question: $\lim_{r \to 1} \pi r^2$
Solution:
To evaluate the limit:
$$\lim_{r \to 1} \pi r^2$$
Direct substitution of $r = 1$ gives:
$$= \pi (1)^2 = \pi$$
Answer: $\pi$
Ex 12.1 - Question 4
Question: $\lim_{x \to 4} \frac{4x + 3}{x - 2}$
Solution:
To find the limit:
$$\lim_{x \to 4} \frac{4x + 3}{x - 2}$$
Since direct substitution of $x = 4$ does not result in an indeterminate form (the denominator $4 - 2 = 2 \neq 0$), we can directly substitute $x = 4$:
$$= \frac{4(4) + 3}{4 - 2}$$
$$= \frac{16 + 3}{2}$$
$$= \frac{19}{2}$$
Answer: $\frac{19}{2}$
Ex 12.1 - Question 5
Question: $\lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1}$
Solution:
To evaluate the limit:
$$\lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1}$$
Substitute $x = -1$ directly into the expression since the denominator $-1 - 1 = -2 \neq 0$:
$$= \frac{(-1)^{10} + (-1)^5 + 1}{-1 - 1}$$
$$= \frac{1 + (-1) + 1}{-2}$$
$$= \frac{1 - 1 + 1}{-2}$$
$$= \frac{1}{-2} = -\frac{1}{2}$$
Answer: $-\frac{1}{2}$
Ex 12.1 - Question 6
Question: $\lim_{x \to 0} \frac{(x + 1)^5 - 1}{x}$
Solution:
We need to evaluate:
$$\lim_{x \to 0} \frac{(x + 1)^5 - 1}{x}$$
Let $x + 1 = y$. As $x \to 0$, $y \to 1$. The expression becomes:
$$\lim_{y \to 1} \frac{y^5 - 1^5}{y - 1}$$
Using the standard limit formula:
$$\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$$
Here, $n = 5$ and $a = 1$:
$$= 5(1)^{5-1} = 5(1)^4 = 5$$
Answer: $5$
Ex 12.1 - Question 7
Question: $\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4}$
Solution:
We need to evaluate:
$$\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4}$$
Direct substitution of $x = 2$ gives the indeterminate form $\frac{0}{0}$. Therefore, we factorize both the numerator and the denominator.
Numerator: $3x^2 - x - 10$
$$= 3x^2 - 6x + 5x - 10 = 3x(x - 2) + 5(x - 2) = (3x + 5)(x - 2)$$
Denominator: $x^2 - 4 = (x - 2)(x + 2)$
Substitute the factors back into the limit:
$$\lim_{x \to 2} \frac{(3x + 5)(x - 2)}{(x - 2)(x + 2)}$$
Cancel out the common term $(x - 2)$ (since $x \neq 2$):
$$= \lim_{x \to 2} \frac{3x + 5}{x + 2}$$
Now, apply direct substitution:
$$= \frac{3(2) + 5}{2 + 2} = \frac{6 + 5}{4} = \frac{11}{4}$$
Answer: $\frac{11}{4}$
Ex 12.1 - Question 8
Question: $\lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3}$
Solution:
Step 1: Direct substitution of $x = 3$ into the expression gives the indeterminate form $\frac{0}{0}$.
Step 2: Factorize both the numerator and the denominator.
Numerator: $x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)$
Denominator: $2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)$
Step 3: Substitute the factorized forms into the limit expression:
$$\lim_{x \to 3} \frac{(x - 3)(x + 3)(x^2 + 9)}{(2x + 1)(x - 3)}$$
Step 4: Cancel the common factor $(x - 3)$ since $x \neq 3$:
$$\lim_{x \to 3} \frac{(x + 3)(x^2 + 9)}{2x + 1}$$
Step 5: Apply the limit by substituting $x = 3$:
$$\frac{(3 + 3)(3^2 + 9)}{2(3) + 1} = \frac{6(9 + 9)}{6 + 1} = \frac{6 \times 18}{7} = \frac{108}{7}$$
Answer: $\frac{108}{7}$
Ex 12.1 - Question 9
Question: $\lim_{x \to 0} \frac{ax + b}{cx + 1}$
Solution:
Step 1: Check for direct substitution. The denominator $cx + 1$ at $x = 0$ becomes $c(0) + 1 = 1 \neq 0$.
Step 2: Since the denominator is non-zero at $x = 0$, we can directly substitute $x = 0$ into the expression:
$$\lim_{x \to 0} \frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0) + 1} = \frac{0 + b}{0 + 1} = b$$
Answer: $b$
Ex 12.1 - Question 10
Question: $\lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$
Solution:
Step 1: Direct substitution of $z = 1$ gives the indeterminate form $\frac{0}{0}$.
Step 2: Use the standard limit formula $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$ or apply substitution.
Let $z = t^6$. As $z \to 1$, $t \to 1$.
Then, $z^{\frac{1}{3}} = (t^6)^{\frac{1}{3}} = t^2$ and $z^{\frac{1}{6}} = (t^6)^{\frac{1}{6}} = t$.
Step 3: Rewrite the limit in terms of $t$:
$$\lim_{t \to 1} \frac{t^2 - 1}{t - 1}$$
Step 4: Factorize the numerator:
$$\lim_{t \to 1} \frac{(t - 1)(t + 1)}{t - 1}$$
Step 5: Cancel $(t - 1)$ and evaluate the limit:
$$\lim_{t \to 1} (t + 1) = 1 + 1 = 2$$
Answer: $2$
Ex 12.1 - Question 11
Question: $\lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a}, a + b + c \neq 0$
Solution:
Step 1: Check for direct substitution of $x = 1$.
Numerator at $x = 1$: $a(1)^2 + b(1) + c = a + b + c$
Denominator at $x = 1$: $c(1)^2 + b(1) + a = a + b + c$
Step 2: Since it is given that $a + b + c \neq 0$, direct substitution yields:
$$\lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a} = \frac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a} = \frac{a + b + c}{a + b + c} = 1$$
Answer: $1$
Ex 12.1 - Question 12
Question: $\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$
Solution:
Given limit:
$$\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$$
First, simplify the numerator by taking the common denominator:
$$\frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2x}$$
Substitute this back into the limit expression:
$$\lim_{x \to -2} \frac{\frac{2 + x}{2x}}{x + 2} = \lim_{x \to -2} \frac{2 + x}{2x(x + 2)}$$
Since $x \to -2$, $x + 2 \neq 0$. We can cancel the common factor $(x + 2)$ from the numerator and the denominator:
$$= \lim_{x \to -2} \frac{1}{2x}$$
Now, substitute $x = -2$:
$$= \frac{1}{2(-2)} = -\frac{1}{4}$$
Answer: $-\frac{1}{4}$
Ex 12.1 - Question 13
Question: $\lim_{x \to 0} \frac{\sin ax}{bx}$
Solution:
Given limit:
$$\lim_{x \to 0} \frac{\sin ax}{bx}$$
Using the standard trigonometric limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we can rearrange the expression:
$$\lim_{x \to 0} \frac{\sin ax}{bx} = \lim_{x \to 0} \left( \frac{\sin ax}{ax} \cdot \frac{a}{b} \right)$$
Since $\frac{a}{b}$ is a constant with respect to $x$, we can factor it out of the limit:
$$= \frac{a}{b} \lim_{x \to 0} \frac{\sin ax}{ax}$$
Let $\theta = ax$. As $x \to 0$, $\theta \to 0$ (since $a \neq 0$). Thus:
$$= \frac{a}{b} \lim_{\theta \to 0} \frac{\sin \theta}{\theta}$$
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$$= \frac{a}{b} \cdot 1 = \frac{a}{b}$$
Answer: $\frac{a}{b}$
Ex 12.1 - Question 14
Question: $\lim_{x \to 0} \frac{\sin ax}{\sin bx}, a, b \neq 0$
Solution:
Given limit:
$$\lim_{x \to 0} \frac{\sin ax}{\sin bx}$$
Divide both the numerator and the denominator by $x$:
$$= \lim_{x \to 0} \frac{\frac{\sin ax}{x}}{\frac{\sin bx}{x}}$$
Multiply and divide the numerator by $a$, and the denominator by $b$, to apply the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$$= \lim_{x \to 0} \frac{a \left(\frac{\sin ax}{ax}\right)}{b \left(\frac{\sin bx}{bx}\right)}$$
Using the quotient law of limits and standard limits ($\lim_{ax \to 0} \frac{\sin ax}{ax} = 1$ and $\lim_{bx \to 0} \frac{\sin bx}{bx} = 1$):
$$= \frac{a \cdot 1}{b \cdot 1} = \frac{a}{b}$$
Answer: $\frac{a}{b}$
Ex 12.1 - Question 15
Question: $\lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)}$
Solution:
Given limit:
$$\lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)}$$
Let $y = \pi - x$.
As $x \to \pi$, $y \to \pi - \pi = 0$.
Substitute $y$ into the limit:
$$\lim_{y \to 0} \frac{\sin y}{\pi y}$$
Factor out the constant $\frac{1}{\pi}$:
$$= \frac{1}{\pi} \lim_{y \to 0} \frac{\sin y}{y}$$
Using the standard limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$:
$$= \frac{1}{\pi} \cdot 1 = \frac{1}{\pi}$$
Answer: $\frac{1}{\pi}$
Ex 12.1 - Question 16
Question: Evaluate $\lim_{x \to 0} \frac{\cos x}{\pi - x}$
Solution: To evaluate the limit: $$\lim_{x \to 0} \frac{\cos x}{\pi - x}$$ Since the denominator $\pi - x$ does not become zero as $x \to 0$ (its limit is $\pi - 0 = \pi \neq 0$), and the function is continuous at $x = 0$, we can directly substitute $x = 0$ using direct substitution method: $$\lim_{x \to 0} \frac{\cos x}{\pi - x} = \frac{\cos(0)}{\pi - 0} = \frac{1}{\pi}$$ Thus, the required limit is $\frac{1}{\pi}$. Answer: $\frac{1}{\pi}$
Ex 12.1 - Question 17
Question: Evaluate $\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}$
Solution: We are given: $$\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}$$ Using trigonometric identity $\cos 2x = 1 - 2\sin^2 x$, we can rewrite the numerator as: $$\cos 2x - 1 = (1 - 2\sin^2 x) - 1 = -2\sin^2 x$$ Substitute this back into the limit expression: $$\lim_{x \to 0} \frac{-2\sin^2 x}{\cos x - 1} = \lim_{x \to 0} \frac{2\sin^2 x}{1 - \cos x}$$ Now, recall the standard trigonometric identity $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$ or use multiplication by the conjugate $(1 + \cos x)$: $$\lim_{x \to 0} \frac{2\sin^2 x}{1 - \cos x} \cdot \frac{1 + \cos x}{1 + \cos x} = \lim_{x \to 0} \frac{2\sin^2 x (1 + \cos x)}{1 - \cos^2 x}$$ Since $1 - \cos^2 x = \sin^2 x$: $$\lim_{x \to 0} \frac{2\sin^2 x (1 + \cos x)}{\sin^2 x} = \lim_{x \to 0} 2(1 + \cos x)$$ Evaluating the limit by substituting $x = 0$: $$= 2(1 + \cos 0) = 2(1 + 1) = 2(2) = 4$$ Thus, the required limit is $4$. Answer: $4$
Ex 12.1 - Question 18
Question: Evaluate $\lim_{x \to 0} \frac{ax + x\cos x}{b \sin x}$
Solution: We are given: $$\lim_{x \to 0} \frac{ax + x\cos x}{b \sin x}$$ Factoring out $x$ from the numerator and separating the terms: $$= \lim_{x \to 0} \frac{x(a + \cos x)}{b \sin x}$$ $$\lim_{x \to 0} \left[ \frac{1}{b} \cdot \frac{x}{\sin x} \cdot (a + \cos x) \right]$$ Using the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$, which implies $\lim_{x \to 0} \frac{x}{\sin x} = 1$: $$= \frac{1}{b} \cdot \left(\lim_{x \to 0} \frac{x}{\sin x}\right) \cdot \left(\lim_{x \to 0} (a + \cos x)\right)$$ $$= \frac{1}{b} \cdot 1 \cdot (a + \cos 0)$$ Since $\cos 0 = 1$: $$= \frac{1}{b} (a + 1) = \frac{a + 1}{b}$$ Thus, the required limit is $\frac{a + 1}{b}$. Answer: $\frac{a + 1}{b}$
Ex 12.1 - Question 19
Question: Evaluate $\lim_{x \to 0} x \sec x$
Solution: We are given: $$\lim_{x \to 0} x \sec x$$ Since $\sec x = \frac{1}{\cos x}$, we can rewrite the limit as: $$\lim_{x \to 0} \frac{x}{\cos x}$$ Applying direct substitution by substituting $x = 0$: $$= \frac{0}{\cos 0} = \frac{0}{1} = 0$$ Thus, the required limit is $0$. Answer: $0$
Ex 12.1 - Question 20
Question: Evaluate $\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}, a, b, a + b \neq 0$
Solution: We are given the limit: $$\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}$$ Divide both the numerator and the denominator by $x$ (since $x \neq 0$ as $x \to 0$): $$= \lim_{x \to 0} \frac{\frac{\sin ax}{x} + b}{a + \frac{\sin bx}{x}}$$ Using the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$: Multiply and divide the terms appropriately: $$= \lim_{x \to 0} \frac{a \left(\frac{\sin ax}{ax}\right) + b}{a + b \left(\frac{\sin bx}{bx}\right)}$$ Since as $x \to 0$, $ax \to 0$ and $bx \to 0$, we have: $$= \frac{a(1) + b}{a + b(1)} = \frac{a + b}{a + b}$$ Given that $a + b \neq 0$, we can cancel the term: $$= 1$$ Answer: $1$
Ex 12.1 - Question 21
Question: $\lim_{x \to 0} (\operatorname{cosec} x - \cot x)$
Solution: We are given the limit: $$\lim_{x \to 0} (\operatorname{cosec} x - \cot x)$$ Rewrite $\operatorname{cosec} x$ and $\cot x$ in terms of $\sin x$ and $\cos x$: $$= \lim_{x \to 0} \left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right)$$ Combine the fractions under a common denominator: $$= \lim_{x \to 0} \frac{1 - \cos x}{\sin x}$$ Use the half-angle formulas $1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)$ and $\sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)$: $$= \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}$$ Simplify the expression: $$= \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \lim_{x \to 0} \tan\left(\frac{x}{2}\right)$$ Substituting $x = 0$: $$= \tan(0) = 0$$ Answer: $0$
Ex 12.1 - Question 22
Question: $\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x - \frac{\pi}{2}}$
Solution: We are given the limit: $$\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x - \frac{\pi}{2}}$$ Let $x - \frac{\pi}{2} = h$. Then as $x \to \frac{\pi}{2}$, $h \to 0$, and $x = h + \frac{\pi}{2}$. Substitute these into the limit: $$= \lim_{h \to 0} \frac{\tan 2\left(h + \frac{\pi}{2}\right)}{h}$$ $$= \lim_{h \to 0} \frac{\tan (2h + \pi)}{h}$$ Using the trigonometric identity $\tan(\theta + \pi) = \tan \theta$: $$= \lim_{h \to 0} \frac{\tan 2h}{h}$$ Multiply and divide by $2$ to use the standard limit $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$: $$= \lim_{h \to 0} 2 \cdot \frac{\tan 2h}{2h} = 2 \cdot 1 = 2$$ Answer: $2$
Ex 12.1 - Question 23
Question: Find $\lim_{x \to 0} f(x)$ and $\lim_{x \to 1} f(x)$, where $f(x) = \begin{cases} 2x + 3, & x \le 0 \\ 3(x + 1), & x > 0 \end{cases}$
Solution: First, let's find $\lim_{x \to 0} f(x)$. We need to check the left-hand limit (LHL) and right-hand limit (RHL) at $x = 0$: $$\text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2x + 3) = 2(0) + 3 = 3$$ $$\text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 3(x + 1) = 3(0 + 1) = 3$$ Since $\text{LHL} = \text{RHL} = 3$, we have: $$\lim_{x \to 0} f(x) = 3$$ Next, let's find $\lim_{x \to 1} f(x)$. Since $1 > 0$, for $x$ in a neighborhood around $1$, the function is defined by $f(x) = 3(x + 1)$. $$\lim_{x \to 1} f(x) = \lim_{x \to 1} 3(x + 1) = 3(1 + 1) = 6$$ Thus, $\lim_{x \to 0} f(x) = 3$ and $\lim_{x \to 1} f(x) = 6$. Answer: $3, 6$
Ex 12.1 - Question 24
Question: Find $\lim_{x \to 1} f(x)$, where $f(x) = \begin{cases} x^2 - 1, & x \le 1 \\ -x^2 - 1, & x > 1 \end{cases}$
Solution: To determine whether $\lim_{x \to 1} f(x)$ exists, we need to evaluate the left-hand limit (LHL) and the right-hand limit (RHL) at $x = 1$. 1. Left-Hand Limit (LHL): $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 - 1)$$ Substituting $x = 1$ into the expression: $$(1)^2 - 1 = 1 - 1 = 0$$ 2. Right-Hand Limit (RHL): $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (-x^2 - 1)$$ Substituting $x = 1$ into the expression: $$-(1)^2 - 1 = -1 - 1 = -2$$ Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$ (i.e., $0 \neq -2$), the limit does not exist at $x = 1$. Answer: Limit does not exist at $x = 1$
Ex 12.1 - Question 25
Question: Evaluate $\lim_{x \to 0} f(x)$, where $f(x) = \begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$
Solution: We evaluate the left-hand limit and right-hand limit of $f(x)$ as $x \to 0$: 1. Left-Hand Limit (LHL): $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{|x|}{x}$$ For $x < 0$, $|x| = -x$. Therefore: $$\lim_{x \to 0^-} \frac{-x}{x} = \lim_{x \to 0^-} (-1) = -1$$ 2. Right-Hand Limit (RHL): $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{|x|}{x}$$ For $x > 0$, $|x| = x$. Therefore: $$\lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} (1) = 1$$ Since $\text{LHL} \neq \text{RHL}$ ($-1 \neq 1$), the limit does not exist at $x = 0$. Answer: Limit does not exist at $x = 0$
Ex 12.1 - Question 26
Question: Find $\lim_{x \to 0} f(x)$, where $f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 0, & x = 0 \end{cases}$
Solution: We evaluate the left-hand limit and right-hand limit of $f(x)$ as $x \to 0$: 1. Left-Hand Limit (LHL): $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|}$$ For $x < 0$, $|x| = -x$. Therefore: $$\lim_{x \to 0^-} \frac{x}{-x} = \lim_{x \to 0^-} (-1) = -1$$ 2. Right-Hand Limit (RHL): $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{|x|}$$ For $x > 0$, $|x| = x$. Therefore: $$\lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} (1) = 1$$ Since $\text{LHL} \neq \text{RHL}$ ($-1 \neq 1$), the limit does not exist at $x = 0$. Answer: Limit does not exist at $x = 0$
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