Limits and Derivatives — NCERT Solutions

Given: $\lim_{x\to 3}(x + 3)$

Concept: For a polynomial function, the limit is found by direct substitution.

Working:
$$\lim_{x\to 3}(x + 3) = 3 + 3 = 6$$

Answer: $6$

Given: $\lim_{x\to \pi}\left(x - \dfrac{22}{7}\right)$

Concept: Direct substitution for a polynomial/linear function.

Working:
$$\lim_{x\to \pi}\left(x - \frac{22}{7}\right) = \pi - \frac{22}{7}$$

Answer: $\pi - \dfrac{22}{7}$

Given: $\lim_{r\to 1}\pi r^2$

Concept: Direct substitution.

Working:
$$\lim_{r\to 1}\pi r^2 = \pi(1)^2 = \pi$$

Answer: $\pi$

Given: $\lim_{x\to 4}\dfrac{4x + 3}{x - 2}$

Concept: Direct substitution (denominator $\neq 0$ at $x = 4$).

Working:
$$\lim_{x\to 4}\frac{4x + 3}{x - 2} = \frac{4(4) + 3}{4 - 2} = \frac{16 + 3}{2} = \frac{19}{2}$$

Answer: $\dfrac{19}{2}$

Given: $\lim_{x\to -1}\dfrac{x^{10} + x^5 + 1}{x - 1}$

Concept: Direct substitution (denominator $\neq 0$ at $x = -1$).

Working:
$$\lim_{x\to -1}\frac{x^{10} + x^5 + 1}{x - 1} = \frac{(-1)^{10} + (-1)^5 + 1}{-1 - 1} = \frac{1 - 1 + 1}{-2} = \frac{1}{-2} = -\frac{1}{2}$$

Answer: $-\dfrac{1}{2}$

Given: $\lim_{x\to 0}\dfrac{(x + 1)^5 - 1}{x}$

Concept: Use the standard limit $\lim_{x\to a}\dfrac{x^n - a^n}{x - a} = na^{n-1}$.

Let $y = x + 1$, so as $x \to 0$, $y \to 1$.

Working:
$$\lim_{x\to 0}\frac{(x+1)^5 - 1}{x} = \lim_{y\to 1}\frac{y^5 - 1^5}{y - 1} = 5 \cdot 1^{5-1} = 5 \times 1 = 5$$

Answer: $5$

Given: $\lim_{x\to 2}\dfrac{3x^2 - x - 10}{x^2 - 4}$

At $x = 2$: numerator $= 12 - 2 - 10 = 0$, denominator $= 0$. So we factorise.

Factorising the numerator:
$3x^2 - x - 10 = (x - 2)(3x + 5)$

Factorising the denominator:
$x^2 - 4 = (x - 2)(x + 2)$

Working:
$$\lim_{x\to 2}\frac{(x-2)(3x+5)}{(x-2)(x+2)} = \lim_{x\to 2}\frac{3x+5}{x+2} = \frac{3(2)+5}{2+2} = \frac{11}{4}$$

Answer: $\dfrac{11}{4}$

Given: $\lim_{x\to 3}\dfrac{x^4 - 81}{2x^2 - 5x - 3}$

At $x = 3$: numerator $= 81 - 81 = 0$, denominator $= 18 - 15 - 3 = 0$. Factorise.

Factorising numerator:
$x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x-3)(x+3)(x^2+9)$

Factorising denominator:
$2x^2 - 5x - 3 = (x - 3)(2x + 1)$

Working:
$$\lim_{x\to 3}\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)} = \lim_{x\to 3}\frac{(x+3)(x^2+9)}{2x+1} = \frac{(6)(18)}{7} = \frac{108}{7}$$

Answer: $\dfrac{108}{7}$

Given: $\lim_{x\to 0}\dfrac{ax + b}{cx + 1}$

Concept: Direct substitution (denominator $= 1 \neq 0$ at $x = 0$).

Working:
$$\lim_{x\to 0}\frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0) + 1} = \frac{b}{1} = b$$

Answer: $b$

Given: $\lim_{z\to 1}\dfrac{z^3 - 1}{z^6 - 1}$

At $z = 1$: both numerator and denominator are $0$. Factorise.

Working:
$$\lim_{z\to 1}\frac{z^3 - 1}{z^6 - 1} = \lim_{z\to 1}\frac{z^3 - 1}{(z^3)^2 - 1^2} = \lim_{z\to 1}\frac{z^3 - 1}{(z^3 - 1)(z^3 + 1)}$$
$$= \lim_{z\to 1}\frac{1}{z^3 + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$

Answer: $\dfrac{1}{2}$

Given: $\lim_{x\to 1}\dfrac{ax^2 + bx + c}{cx^2 + bx + a}$, where $a + b + c \neq 0$.

Concept: At $x = 1$, numerator $= a + b + c \neq 0$ and denominator $= c + b + a = a + b + c \neq 0$. So direct substitution applies.

Working:
$$\lim_{x\to 1}\frac{ax^2 + bx + c}{cx^2 + bx + a} = \frac{a(1) + b(1) + c}{c(1) + b(1) + a} = \frac{a + b + c}{a + b + c} = 1$$

Answer: $1$

Given: $\lim_{x\to -2}\dfrac{\dfrac{1}{x} + \dfrac{1}{2}}{x + 2}$

At $x = -2$: numerator $= -\frac{1}{2} + \frac{1}{2} = 0$, denominator $= 0$. Simplify.

Working:
$$\frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \frac{\frac{2 + x}{2x}}{x + 2} = \frac{x + 2}{2x(x + 2)} = \frac{1}{2x}$$

$$\lim_{x\to -2}\frac{1}{2x} = \frac{1}{2(-2)} = -\frac{1}{4}$$

Answer: $-\dfrac{1}{4}$

Given: $\lim_{x\to 0}\dfrac{\sin ax}{bx}$

Concept: Use the standard limit $\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta} = 1$.

Working:
$$\lim_{x\to 0}\frac{\sin ax}{bx} = \lim_{x\to 0}\frac{\sin ax}{ax} \cdot \frac{ax}{bx} = \lim_{x\to 0}\frac{\sin ax}{ax} \cdot \frac{a}{b}$$
$$= 1 \cdot \frac{a}{b} = \frac{a}{b}$$

Answer: $\dfrac{a}{b}$

Given: $\lim_{x\to 0}\dfrac{\sin ax}{\sin bx}$, $a, b \neq 0$.

Concept: Use $\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta} = 1$.

Working:
$$\lim_{x\to 0}\frac{\sin ax}{\sin bx} = \lim_{x\to 0}\frac{\dfrac{\sin ax}{ax} \cdot ax}{\dfrac{\sin bx}{bx} \cdot bx} = \frac{\lim_{x\to 0}\dfrac{\sin ax}{ax}}{\lim_{x\to 0}\dfrac{\sin bx}{bx}} \cdot \frac{a}{b} = \frac{1}{1} \cdot \frac{a}{b} = \frac{a}{b}$$

Answer: $\dfrac{a}{b}$

Given: $\lim_{x\to \pi}\dfrac{\sin(\pi - x)}{\pi(\pi - x)}$

Concept: Let $y = \pi - x$. As $x \to \pi$, $y \to 0$.

Working:
$$\lim_{x\to \pi}\frac{\sin(\pi - x)}{\pi(\pi - x)} = \lim_{y\to 0}\frac{\sin y}{\pi y} = \frac{1}{\pi}\lim_{y\to 0}\frac{\sin y}{y} = \frac{1}{\pi} \cdot 1 = \frac{1}{\pi}$$

Answer: $\dfrac{1}{\pi}$

Given: $\lim_{x\to 0}\dfrac{\cos x}{\pi - x}$

Concept: Direct substitution (denominator $= \pi \neq 0$ at $x = 0$).

Working:
$$\lim_{x\to 0}\frac{\cos x}{\pi - x} = \frac{\cos 0}{\pi - 0} = \frac{1}{\pi}$$

Answer: $\dfrac{1}{\pi}$

Given: $\lim_{x\to 0}\dfrac{\cos 2x - 1}{\cos x - 1}$

Concept: Use $1 - \cos\theta = 2\sin^2\dfrac{\theta}{2}$.

Working:
$$\cos 2x - 1 = -2\sin^2 x, \quad \cos x - 1 = -2\sin^2\frac{x}{2}$$

$$\lim_{x\to 0}\frac{-2\sin^2 x}{-2\sin^2\frac{x}{2}} = \lim_{x\to 0}\frac{\sin^2 x}{\sin^2\frac{x}{2}}$$

Using $\sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}$:
$$= \lim_{x\to 0}\frac{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}} = \lim_{x\to 0}4\cos^2\frac{x}{2} = 4\cos^2 0 = 4 \times 1 = 4$$

Answer: $4$

Given: $\lim_{x\to 0}\dfrac{ax + x\cos x}{b\sin x}$

Working:
$$\lim_{x\to 0}\frac{x(a + \cos x)}{b\sin x} = \lim_{x\to 0}\frac{x}{\sin x} \cdot \frac{a + \cos x}{b}$$
$$= \lim_{x\to 0}\frac{1}{\dfrac{\sin x}{x}} \cdot \frac{a + \cos x}{b} = \frac{1}{1} \cdot \frac{a + \cos 0}{b} = \frac{a + 1}{b}$$

Answer: $\dfrac{a + 1}{b}$

Given: $\lim_{x\to 0}x\sec x$

Concept: Direct substitution.

Working:
$$\lim_{x\to 0}x\sec x = \lim_{x\to 0}\frac{x}{\cos x} = \frac{0}{\cos 0} = \frac{0}{1} = 0$$

Answer: $0$

Given: $\lim_{x\to 0}\dfrac{\sin ax + bx}{ax + \sin bx}$

Concept: Divide numerator and denominator by $x$.

Working:
$$\lim_{x\to 0}\frac{\dfrac{\sin ax}{x} + b}{a + \dfrac{\sin bx}{x}} = \lim_{x\to 0}\frac{a\cdot\dfrac{\sin ax}{ax} + b}{a + b\cdot\dfrac{\sin bx}{bx}}$$
$$= \frac{a \cdot 1 + b}{a + b \cdot 1} = \frac{a + b}{a + b} = 1$$

Answer: $1$

Given: $\lim_{x\to 0}(\csc x - \cot x)$

Working:
$$\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right) = \lim_{x\to 0}\frac{1 - \cos x}{\sin x}$$

Multiply numerator and denominator by $x$:
$$= \lim_{x\to 0}\frac{1 - \cos x}{x} \cdot \frac{x}{\sin x} = 0 \cdot 1 = 0$$

(Using standard limits: $\lim_{x\to 0}\dfrac{1-\cos x}{x} = 0$ and $\lim_{x\to 0}\dfrac{x}{\sin x} = 1$.)

Answer: $0$

Given: $\lim_{x\to \frac{\pi}{2}}\dfrac{\tan 2x}{x - \dfrac{\pi}{2}}$

Concept: Let $y = x - \dfrac{\pi}{2}$, so as $x \to \dfrac{\pi}{2}$, $y \to 0$, and $x = y + \dfrac{\pi}{2}$.

Working:
$$\tan 2x = \tan 2\left(y + \frac{\pi}{2}\right) = \tan(2y + \pi) = \tan 2y$$

$$\lim_{y\to 0}\frac{\tan 2y}{y} = \lim_{y\to 0}\frac{\sin 2y}{y \cos 2y} = \lim_{y\to 0}\frac{\sin 2y}{2y} \cdot \frac{2}{\cos 2y} = 1 \cdot \frac{2}{1} = 2$$

Answer: $2$

Given: $f(x) = \begin{cases} 2x + 3, & x\leq 0 \\ 3(x + 1), & x > 0 \end{cases}$

Finding $\lim_{x\to 0}f(x)$:

Left-hand limit (LHL):
$$\lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(2x + 3) = 2(0) + 3 = 3$$

Right-hand limit (RHL):
$$\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}3(x + 1) = 3(0 + 1) = 3$$

Since LHL $=$ RHL $= 3$:
$$\lim_{x\to 0}f(x) = 3$$

Finding $\lim_{x\to 1}f(x)$:

For $x$ near $1$ (both sides), $x > 0$, so $f(x) = 3(x+1)$.

LHL:
$$\lim_{x\to 1^-}f(x) = 3(1 + 1) = 6$$

RHL:
$$\lim_{x\to 1^+}f(x) = 3(1 + 1) = 6$$

Since LHL $=$ RHL $= 6$:
$$\lim_{x\to 1}f(x) = 6$$

Given: $f(x) = \begin{cases} x^2 - 1, & x \leq 1 \\ -x^2 - 1, & x > 1 \end{cases}$

LHL:
$$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}(x^2 - 1) = 1 - 1 = 0$$

RHL:
$$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}(-x^2 - 1) = -1 - 1 = -2$$

Since LHL $\neq$ RHL, $\lim_{x\to 1}f(x)$ does not exist.

Given: $f(x) = \begin{cases} \dfrac{|x|}{x}, & x\neq 0\\ 0, & x = 0 \end{cases}$

For $x > 0$: $|x| = x$, so $f(x) = \dfrac{x}{x} = 1$.

For $x < 0$: $|x| = -x$, so $f(x) = \dfrac{-x}{x} = -1$.

LHL:
$$\lim_{x\to 0^-}f(x) = -1$$

RHL:
$$\lim_{x\to 0^+}f(x) = 1$$

Since LHL $\neq$ RHL, $\lim_{x\to 0}f(x)$ does not exist.

Given: $f(x) = \begin{cases} \dfrac{x}{|x|}, & x\neq 0\\ 0, & x = 0 \end{cases}$

For $x > 0$: $f(x) = \dfrac{x}{x} = 1$.

For $x < 0$: $f(x) = \dfrac{x}{-x} = -1$.

LHL:
$$\lim_{x\to 0^-}f(x) = -1$$

RHL:
$$\lim_{x\to 0^+}f(x) = 1$$

Since LHL $\neq$ RHL, $\lim_{x\to 0}f(x)$ does not exist.

Given: $f(x) = |x| - 5$

For $x$ near $5$ (both sides), $x > 0$, so $|x| = x$.

LHL:
$$\lim_{x\to 5^-}f(x) = \lim_{x\to 5^-}(x - 5) = 5 - 5 = 0$$

RHL:
$$\lim_{x\to 5^+}f(x) = \lim_{x\to 5^+}(x - 5) = 5 - 5 = 0$$

Since LHL $=$ RHL $= 0$:
$$\lim_{x\to 5}f(x) = 0$$

Given: $f(1) = 4$ and $\lim_{x\to 1}f(x) = f(1) = 4$.

For the limit to exist, LHL must equal RHL.

LHL:
$$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}(a + bx) = a + b$$

RHL:
$$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}(b - ax) = b - a$$

For the limit to exist:
$$a + b = b - a \implies 2a = 0 \implies a = 0$$

For the limit to equal $f(1) = 4$:
$$a + b = 4 \implies 0 + b = 4 \implies b = 4$$

Answer: $a = 0$ and $b = 4$.

Given: $f(x) = (x - a_1)(x - a_2)\cdots(x - a_n)$

Finding $\lim_{x\to a_1}f(x)$:

Since $f(x)$ is a polynomial, the limit equals the value at $x = a_1$:
$$\lim_{x\to a_1}f(x) = (a_1 - a_1)(a_1 - a_2)\cdots(a_1 - a_n) = 0$$

Finding $\lim_{x\to a}f(x)$ for $a \neq a_1, a_2, \ldots, a_n$:

Again by direct substitution:
$$\lim_{x\to a}f(x) = (a - a_1)(a - a_2)\cdots(a - a_n)$$

This is a non-zero finite value since $a$ is different from all $a_i$.

Given: $f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases}$

Case 1: $a = 0$

LHL: $\lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(|x| + 1) = 0 + 1 = 1$

RHL: $\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}(|x| - 1) = 0 - 1 = -1$

LHL $\neq$ RHL, so the limit does not exist at $a = 0$.

Case 2: $a < 0$

For $x$ near $a < 0$, $f(x) = |x| + 1 = -x + 1$ (since $x < 0$).

$$\lim_{x\to a}f(x) = -a + 1$$

LHL $=$ RHL, so the limit exists for all $a < 0$.

Case 3: $a > 0$

For $x$ near $a > 0$, $f(x) = |x| - 1 = x - 1$.

$$\lim_{x\to a}f(x) = a - 1$$

LHL $=$ RHL, so the limit exists for all $a > 0$.

Conclusion: $\lim_{x\to a}f(x)$ exists for all $a \neq 0$, i.e., for all $a \in (-\infty, 0) \cup (0, \infty)$.

Given: $\lim_{x\to 1}\dfrac{f(x) - 2}{x^2 - 1} = \pi$

Working:

As $x \to 1$, the denominator $x^2 - 1 \to 0$. For the limit to be finite (equal to $\pi$), the numerator must also $\to 0$.

Therefore:
$$\lim_{x\to 1}[f(x) - 2] = \lim_{x\to 1}\frac{f(x)-2}{x^2-1} \cdot \lim_{x\to 1}(x^2-1) = \pi \times 0 = 0$$

$$\lim_{x\to 1}f(x) - 2 = 0$$

$$\lim_{x\to 1}f(x) = 2$$

Answer: $\lim_{x\to 1}f(x) = 2$

Given: $f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ nx^3 + m, & x > 1 \end{cases}$

Condition for $\lim_{x\to 0}f(x)$ to exist:

LHL: $\lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(mx^2 + n) = n$

RHL: $\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}(nx + m) = m$

For limit to exist: LHL $=$ RHL $\Rightarrow$ $n = m$.

Condition for $\lim_{x\to 1}f(x)$ to exist:

LHL: $\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}(nx + m) = n + m$

RHL: $\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}(nx^3 + m) = n + m$

LHL $=$ RHL $= n + m$ for all values of $m$ and $n$. So $\lim_{x\to 1}f(x)$ exists for all integers $m$ and $n$.

Conclusion: Both limits exist when $m = n$ (where $m$ and $n$ are any equal integers).

Given: $f(x) = x^2 - 2$

Formula: $f'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$

Working:
$$f'(x) = \lim_{h\to 0}\frac{(x+h)^2 - 2 - (x^2 - 2)}{h} = \lim_{h\to 0}\frac{x^2 + 2xh + h^2 - x^2}{h}$$
$$= \lim_{h\to 0}\frac{2xh + h^2}{h} = \lim_{h\to 0}(2x + h) = 2x$$

At $x = 10$:
$$f'(10) = 2(10) = 20$$

Answer: $20$

Given: $f(x) = x$

Working:
$$f'(x) = \lim_{h\to 0}\frac{(x+h) - x}{h} = \lim_{h\to 0}\frac{h}{h} = 1$$

At $x = 1$: $f'(1) = 1$

Answer: $1$

Given: $f(x) = 99x$

Working:
$$f'(x) = \lim_{h\to 0}\frac{99(x+h) - 99x}{h} = \lim_{h\to 0}\frac{99h}{h} = 99$$

At $x = 100$: $f'(100) = 99$

Answer: $99$

(i) $f(x) = x^3 - 27$

$$f'(x) = \lim_{h\to 0}\frac{(x+h)^3 - 27 - (x^3 - 27)}{h}$$
$$= \lim_{h\to 0}\frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}$$
$$= \lim_{h\to 0}(3x^2 + 3xh + h^2) = 3x^2$$

Answer: $f'(x) = 3x^2$