NCERT Class 11 Mathematics — Limits and Derivatives, Exercise 12.1. All 32 questions solved.

Evaluating a limit follows a fixed decision procedure.

1. Substitute. If the result is a definite number, that is the limit.
2. If you get $\tfrac00$, the numerator and denominator share a factor that vanishes. Cancel it and substitute again.
3. For trigonometric limits, reduce to the two standard results

$$\lim_{x\to0}\frac{\sin x}{x} = 1, \qquad \lim_{x\to0}\frac{\tan x}{x} = 1$$

4. For a piecewise function, compute the left-hand and right-hand limits separately. The limit exists only if they agree.

The standard algebraic limit is also worth having to hand:

$$\lim_{x\to a}\frac{x^n – a^n}{x – a} = na^{n-1}$$

Key insight. $\tfrac00$ is not an answer — it is a signal that a common factor is hiding. Getting it means the work has not started. And the value of the function at the point is irrelevant to the limit: in question 25 the function is defined to be $0$ at $x = 0$, yet the limit fails to exist there, because the two one-sided limits are $-1$ and $1$.

Evaluate the following limits in questions 1 to 22.

Question 1

$\displaystyle\lim_{x\to3}\,(x+3)$

Solution. A polynomial is continuous, so substitute directly:

$$3 + 3 = 6$$

Question 2

$\displaystyle\lim_{x\to\pi}\left(x – \frac{22}{7}\right)$

Solution. Substituting $x = \pi$:

$$\pi – \frac{22}{7}$$

Note this is not zero — $\tfrac{22}{7}$ is only an approximation to $\pi$, and the answer must be left exact.

Question 3

$\displaystyle\lim_{r\to1}\pi r^2$

Solution. $\pi$ is a constant, so

$$\pi(1)^2 = \pi$$

Question 4

$\displaystyle\lim_{x\to4}\frac{4x+3}{x-2}$

Solution. The denominator at $x = 4$ is $2 \ne 0$, so substitution is valid:

$$\frac{16+3}{4-2} = \frac{19}{2}$$

Question 5

$\displaystyle\lim_{x\to-1}\frac{x^{10} + x^5 + 1}{x-1}$

Solution. At $x = -1$ the denominator is $-2 \ne 0$, so substitute. Since $(-1)^{10} = 1$ and $(-1)^5 = -1$:

$$\frac{1 – 1 + 1}{-1 – 1} = \frac{1}{-2} = -\frac12$$

Question 6

$\displaystyle\lim_{x\to0}\frac{(x+1)^5 – 1}{x}$

Solution. Substitution gives $\tfrac00$, so expand the numerator:

$$(x+1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1$$

$$(x+1)^5 – 1 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x = x\left(x^4 + 5x^3 + 10x^2 + 10x + 5\right)$$

Cancelling the $x$:

$$\lim_{x\to0}\left(x^4 + 5x^3 + 10x^2 + 10x + 5\right) = 5$$

(Alternatively, substituting $t = x+1$ turns this into $\lim_{t\to1}\tfrac{t^5-1}{t-1} = 5(1)^4 = 5$.)

Question 7

$\displaystyle\lim_{x\to2}\frac{3x^2 – x – 10}{x^2 – 4}$

Solution. Both parts vanish at $x = 2$, so factorise:

$$3x^2 – x – 10 = (x-2)(3x+5), \qquad x^2 – 4 = (x-2)(x+2)$$

Cancelling $(x-2)$:

$$\lim_{x\to2}\frac{3x+5}{x+2} = \frac{11}{4}$$

Question 8

$\displaystyle\lim_{x\to3}\frac{x^4 – 81}{2x^2 – 5x – 3}$

Solution. Both vanish at $x = 3$. Factorise each:

$$x^4 – 81 = \left(x^2 – 9\right)\left(x^2 + 9\right) = (x-3)(x+3)\left(x^2+9\right)$$

$$2x^2 – 5x – 3 = (x – 3)(2x + 1)$$

Cancelling $(x-3)$:

$$\lim_{x\to3}\frac{(x+3)\left(x^2+9\right)}{2x+1} = \frac{6 \times 18}{7} = \frac{108}{7}$$

Question 9

$\displaystyle\lim_{x\to0}\frac{ax+b}{cx+1}$

Solution. The denominator tends to $1 \ne 0$, so substitute:

$$\frac{0 + b}{0 + 1} = b$$

Question 10

$\displaystyle\lim_{z\to1}\frac{z^{1/3} – 1}{z^{1/6} – 1}$

Solution. Substituting gives $\tfrac00$. Put $t = z^{1/6}$, so that $z^{1/3} = t^2$ and $t \to 1$ as $z \to 1$:

$$\lim_{t\to1}\frac{t^2 – 1}{t – 1} = \lim_{t\to1}\frac{(t-1)(t+1)}{t-1} = \lim_{t\to1}(t+1) = 2$$

Choosing the substitution as the lowest power present — here the sixth root — is what makes both exponents whole numbers.

Question 11

$\displaystyle\lim_{x\to1}\frac{ax^2 + bx + c}{cx^2 + bx + a}$,   $a + b + c \ne 0$

Solution. At $x = 1$ the numerator is $a + b + c$ and the denominator is $c + b + a$ — the same quantity. Since it is non-zero by hypothesis:

$$\frac{a+b+c}{a+b+c} = 1$$

The condition $a+b+c \ne 0$ is exactly what makes substitution legitimate.

Question 12

$\displaystyle\lim_{x\to-2}\frac{\frac1x + \frac12}{x+2}$

Solution. Combine the numerator over a common denominator first:

$$\frac1x + \frac12 = \frac{2 + x}{2x}$$

So the whole expression is

$$\frac{\frac{x+2}{2x}}{x+2} = \frac{1}{2x}$$

for $x \ne -2$. Now substitute:

$$\frac{1}{2(-2)} = -\frac14$$

Question 13

$\displaystyle\lim_{x\to0}\frac{\sin ax}{bx}$

Solution. Manufacture the standard form by multiplying above and below by $a$:

$$\frac{\sin ax}{bx} = \frac ab \cdot \frac{\sin ax}{ax}$$

As $x \to 0$, so does $ax$, and $\tfrac{\sin ax}{ax} \to 1$:

$$\frac ab$$

Question 14

$\displaystyle\lim_{x\to0}\frac{\sin ax}{\sin bx}$,   $a, b \ne 0$

Solution. Divide numerator and denominator by $x$ and arrange both into standard form:

$$\frac{\sin ax}{\sin bx} = \frac{\frac{\sin ax}{ax} \cdot ax}{\frac{\sin bx}{bx} \cdot bx} = \frac ab \cdot \frac{\frac{\sin ax}{ax}}{\frac{\sin bx}{bx}}$$

Both fractions tend to $1$:

$$\frac ab$$

Question 15

$\displaystyle\lim_{x\to\pi}\frac{\sin(\pi – x)}{\pi(\pi – x)}$

Solution. Substitute $t = \pi – x$, so $t \to 0$:

$$\lim_{t\to0}\frac{\sin t}{\pi t} = \frac{1}{\pi}\lim_{t\to0}\frac{\sin t}{t} = \frac{1}{\pi}$$

Question 16

$\displaystyle\lim_{x\to0}\frac{\cos x}{\pi – x}$

Solution. No indeterminate form here — the denominator tends to $\pi \ne 0$:

$$\frac{\cos 0}{\pi – 0} = \frac{1}{\pi}$$

This question sits beside question 15 to make the point that a limit involving trigonometric functions is not automatically a $\tfrac00$ problem.

Question 17

$\displaystyle\lim_{x\to0}\frac{\cos 2x – 1}{\cos x – 1}$

Solution. Use $\cos\theta – 1 = -2\sin^2\tfrac\theta2$ on both:

$$\frac{-2\sin^2 x}{-2\sin^2\frac x2} = \frac{\sin^2 x}{\sin^2\frac x2}$$

Divide top and bottom by $x^2$ and arrange into standard forms:

$$= \frac{\left(\frac{\sin x}{x}\right)^2}{\left(\frac{\sin(x/2)}{x/2}\right)^2 \cdot \frac14}$$

As $x \to 0$ both bracketed fractions tend to $1$:

$$\frac{1}{1/4} = 4$$

Question 18

$\displaystyle\lim_{x\to0}\frac{ax + x\cos x}{b\sin x}$

Solution. Factor $x$ out of the numerator and divide through by $x$:

$$\frac{x(a + \cos x)}{b\sin x} = \frac{a + \cos x}{b} \cdot \frac{x}{\sin x}$$

As $x \to 0$: $\cos x \to 1$ and $\tfrac{x}{\sin x} \to 1$:

$$\frac{a+1}{b}$$

Question 19

$\displaystyle\lim_{x\to0}\,x\sec x$

Solution. $\sec x = \tfrac{1}{\cos x} \to 1$ as $x \to 0$, so

$$0 \times 1 = 0$$

Question 20

$\displaystyle\lim_{x\to0}\frac{\sin ax + bx}{ax + \sin bx}$,   $a, b, a+b \ne 0$

Solution. Divide every term by $x$:

$$\frac{\frac{\sin ax}{x} + b}{a + \frac{\sin bx}{x}} = \frac{a \cdot \frac{\sin ax}{ax} + b}{a + b \cdot \frac{\sin bx}{bx}}$$

As $x \to 0$ both standard fractions tend to $1$:

$$\frac{a + b}{a + b} = 1$$

The condition $a + b \ne 0$ keeps the denominator alive.

Question 21

$\displaystyle\lim_{x\to0}\left(\operatorname{cosec} x – \cot x\right)$

Solution. This is an $\infty – \infty$ form, so combine into a single fraction:

$$\frac{1}{\sin x} – \frac{\cos x}{\sin x} = \frac{1 – \cos x}{\sin x}$$

Now use $1 – \cos x = 2\sin^2\tfrac x2$ and $\sin x = 2\sin\tfrac x2\cos\tfrac x2$:

$$= \frac{2\sin^2\frac x2}{2\sin\frac x2\cos\frac x2} = \tan\frac x2$$

As $x \to 0$, $\tan\tfrac x2 \to 0$:

$$0$$

Question 22

$\displaystyle\lim_{x\to\frac\pi2}\frac{\tan 2x}{x – \frac\pi2}$

Solution. Substitute $t = x – \tfrac\pi2$, so $t \to 0$ and $2x = \pi + 2t$. Since $\tan$ has period $\pi$:

$$\tan 2x = \tan(\pi + 2t) = \tan 2t$$

So the limit becomes

$$\lim_{t\to0}\frac{\tan 2t}{t} = \lim_{t\to0}\ 2 \cdot \frac{\tan 2t}{2t} = 2$$

Question 23

Find $\displaystyle\lim_{x\to0}f(x)$ and $\displaystyle\lim_{x\to1}f(x)$, where

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

Solution.

At $x = 0$ — the branch point, so both sides must be checked.

$$\lim_{x\to0^-}f(x) = 2(0) + 3 = 3, \qquad \lim_{x\to0^+}f(x) = 3(0+1) = 3$$

They agree, so $\displaystyle\lim_{x\to0}f(x) = 3$.

At $x = 1$ — well inside the region $x > 0$, so only the second branch applies and no one-sided work is needed:

$$\lim_{x\to1}f(x) = 3(1+1) = 6$$

Question 24

Find $\displaystyle\lim_{x\to1}f(x)$, where

$$f(x) = \begin{cases} x^2 – 1, & x \le 1 \\ -x^2 – 1, & x > 1 \end{cases}$$

Solution.

$$\lim_{x\to1^-}f(x) = 1^2 – 1 = 0$$

$$\lim_{x\to1^+}f(x) = -1^2 – 1 = -2$$

The two one-sided limits differ, so the limit does not exist at $x = 1$.

Question 25

Evaluate $\displaystyle\lim_{x\to0}f(x)$, where

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

Solution. For $x < 0$, $|x| = -x$, so $f(x) = -1$. For $x > 0$, $|x| = x$, so $f(x) = 1$.

$$\lim_{x\to0^-}f(x) = -1, \qquad \lim_{x\to0^+}f(x) = 1$$

These differ, so the limit does not exist. Note that $f(0) = 0$ is irrelevant — a limit never consults the value at the point itself.

Question 26

Find $\displaystyle\lim_{x\to0}f(x)$, where

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

Solution. For $x \ne 0$, $\dfrac{x}{|x|}$ is the reciprocal of $\dfrac{|x|}{x}$, and both equal $\pm1$, so the function is identical to question 25’s:

$$\lim_{x\to0^-}f(x) = -1, \qquad \lim_{x\to0^+}f(x) = 1$$

The limit does not exist.

Question 27

Find $\displaystyle\lim_{x\to5}f(x)$, where $f(x) = |x| – 5$.

Solution. Near $x = 5$ every value of $x$ is positive, so $|x| = x$ and the function is simply $x – 5$ there — no modulus difficulty arises because the branch point $x = 0$ is far away.

$$\lim_{x\to5}(x – 5) = 0$$

Question 28

Suppose

$$f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b – ax, & x > 1 \end{cases}$$

and if $\displaystyle\lim_{x\to1}f(x) = f(1)$, what are the possible values of $a$ and $b$?

Solution. For the limit to exist and equal $f(1) = 4$, both one-sided limits must equal $4$.

Left:

$$\lim_{x\to1^-}f(x) = a + b = 4 \qquad (1)$$

Right:

$$\lim_{x\to1^+}f(x) = b – a = 4 \qquad (2)$$

Adding (1) and (2): $2b = 8$, so $b = 4$. Substituting back into (1): $a = 0$.

Question 29

Let $a_1, a_2, \ldots, a_n$ be fixed real numbers and define a function

$$f(x) = (x – a_1)(x – a_2)\cdots(x – a_n)$$

What is $\displaystyle\lim_{x\to a_1}f(x)$? For some $a \ne a_1, a_2, \ldots, a_n$, compute $\displaystyle\lim_{x\to a}f(x)$.

Solution. $f$ is a polynomial, hence continuous everywhere, so every limit is found by substitution.

At $x = a_1$: the first factor becomes $a_1 – a_1 = 0$, and a product with a zero factor is zero:

$$\lim_{x\to a_1}f(x) = 0$$

At a general $a$:

$$\lim_{x\to a}f(x) = (a – a_1)(a – a_2)\cdots(a – a_n)$$

The condition $a \ne a_i$ makes this product non-zero, but the formula would hold either way.

Question 30

If

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

for what value(s) of $a$ does $\displaystyle\lim_{x\to a}f(x)$ exist?

Solution. Away from $x = 0$ the function is given by a single continuous formula, so the only point in question is the join.

For $a < 0$: near $a$, $f(x) = -x + 1$, continuous, so the limit exists and equals $1 – a$.

For $a > 0$: near $a$, $f(x) = x – 1$, continuous, so the limit exists and equals $a – 1$.

At $a = 0$:

$$\lim_{x\to0^-}f(x) = 0 + 1 = 1, \qquad \lim_{x\to0^+}f(x) = 0 – 1 = -1$$

These differ, so the limit does not exist there.

Question 31

If the function $f(x)$ satisfies $\displaystyle\lim_{x\to1}\frac{f(x) – 2}{x^2 – 1} = \pi$, evaluate $\displaystyle\lim_{x\to1}f(x)$.

Solution. The denominator $x^2 – 1$ tends to $0$ as $x \to 1$. If the numerator tended to any non-zero value, the quotient would grow without bound rather than approach the finite value $\pi$. So the numerator must tend to $0$:

$$\lim_{x\to1}\left(f(x) – 2\right) = 0 \quad\Longrightarrow\quad \lim_{x\to1}f(x) = 2$$

More formally,

$$\lim_{x\to1}\left(f(x) – 2\right) = \lim_{x\to1}\left[\frac{f(x)-2}{x^2-1} \times \left(x^2-1\right)\right] = \pi \times 0 = 0$$

Question 32

If

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

for what integers $m$ and $n$ do both $\displaystyle\lim_{x\to0}f(x)$ and $\displaystyle\lim_{x\to1}f(x)$ exist?

Solution. There are two join points; treat them separately.

At $x = 0$:

$$\lim_{x\to0^-}f(x) = m(0)^2 + n = n$$ $$\lim_{x\to0^+}f(x) = n(0) + m = m$$

The limit exists only if these agree:

$$m = n$$

At $x = 1$:

$$\lim_{x\to1^-}f(x) = n(1) + m = m + n$$ $$\lim_{x\to1^+}f(x) = n(1)^3 + m = m + n$$

They agree automatically, whatever $m$ and $n$ are.

So the constraint comes entirely from $x = 0$.

Common mistakes

• Treating $\tfrac00$ as an answer. It means a common factor is present, not that the limit is $0$ or undefined. Questions 6 to 8, 10 and 12 all begin this way.
• Question 2, writing the answer as $0$. $\pi \ne \tfrac{22}{7}$; the fraction is only an approximation, so the answer stays exact.
• Question 5, assuming it is indeterminate. At $x = -1$ the denominator is $-2$, so substitution works straight away. Checking the denominator before doing anything clever saves the effort.
• Question 13, answering $1$. $\lim\tfrac{\sin ax}{x} = a$, not $1$. The standard form needs the same expression above and below, which is why the $a$ must be inserted deliberately.
• Questions 15 and 22, not substituting. When $x$ tends to something other than $0$, shift the variable so it does — that is what turns these into the standard forms.
• Questions 24 to 26, giving a single value. A limit exists only when both one-sided limits agree. Where they do not, “does not exist” is the answer, and it must be justified by exhibiting both.
• Question 25, letting $f(0) = 0$ decide the limit. The value at the point plays no part; only the behaviour on either side does.
• Question 32, checking only one join point. Both $x = 0$ and $x = 1$ must be examined; the second turns out to impose nothing, but that has to be shown rather than assumed.

Practise next

• Exercise 12.2 — derivatives from first principles, built on the limit $\lim_{h\to0}\tfrac{f(x+h)-f(x)}{h}$.
• Miscellaneous Exercise on Chapter 12 — a long mixed set of derivatives, using the product and quotient rules.