NCERT Exemplar Solutions Class 11 Maths Limits and Derivatives — Long Answer and Objective Type Q47–64

NCERT Exemplar Problems, Class 11 Mathematics — Limits and Derivatives, Long Answer Type questions 47 to 53 and Objective Type questions 54 to 64. All 18 questions solved.

Questions 47 to 53 are the harder limits: two need a limit-as-derivative recognition (47) or a long trigonometric simplification (48 to 50), one asks you to show that a limit does not exist (51), and two find a constant that makes a limit exist (52, 53). Questions 54 to 64 are objective: standard limits (54 to 60, 62), one limit with a greatest-integer function (63), the modulus-sine limit (64), and one with a flaw (61).

Question 61 has no correct option among the four printed, and the key’s (D) is wrong; Question 53’s inequalities are misprinted in the book. Both are explained where they occur.

Key insight. For a limit that does not exist, or a function defined in pieces, always check the left-hand and right-hand limits separately. They must be equal for the limit to exist; if they differ, the limit does not exist.

Evaluate each of the following limits in Questions 47 to 53.

Question 47

$\displaystyle\lim_{y\to0}\frac{(x+y)\sec(x+y)-x\sec x}{y}$

Solution. Let $f(x)=x\sec x$. Then the expression is $\dfrac{f(x+y)-f(x)}{y}$, which is exactly the definition of $f'(x)$ with $y$ playing the role of $h$. So the limit is the derivative, by the product rule:

$$\frac{d}{dx}(x\sec x)=\sec x+x\sec x\tan x=\sec x\,(1+x\tan x)$$

Question 48

$\displaystyle\lim_{x\to0}\frac{\sin(\alpha+\beta)x+\sin(\alpha-\beta)x+\sin2\alpha x}{\cos2\beta x-\cos2\alpha x}\cdot x$

Solution. Turn the numerator into a product. By the sum formula, $\sin(\alpha+\beta)x+\sin(\alpha-\beta)x=2\sin\alpha x\cos\beta x$, and $\sin2\alpha x=2\sin\alpha x\cos\alpha x$, so the numerator is $2\sin\alpha x\,(\cos\beta x+\cos\alpha x)$.

For the denominator, $\cos2\beta x-\cos2\alpha x=2\cos^2\beta x-2\cos^2\alpha x=2(\cos\beta x-\cos\alpha x)(\cos\beta x+\cos\alpha x)$.

The factor $\cos\beta x+\cos\alpha x$ cancels (it tends to $2$, not $0$), leaving

$$\lim_{x\to0}\frac{x\sin\alpha x}{\cos\beta x-\cos\alpha x}.$$

Now $\cos\beta x-\cos\alpha x=(1-\cos\alpha x)-(1-\cos\beta x)$, and after dividing by $x^2$ it tends to $\frac{\alpha^2}2-\frac{\beta^2}2$. The numerator over $x^2$ is $\dfrac{\sin\alpha x}{x}\to\alpha$. So

$$\frac{\alpha}{\frac{\alpha^2-\beta^2}2}=\frac{2\alpha}{\alpha^2-\beta^2}.$$

Question 49

$\displaystyle\lim_{x\to\frac\pi4}\frac{\tan^3x-\tan x}{\cos\left(x+\frac\pi4\right)}$

Solution. Factorise the numerator: $\tan^3x-\tan x=\tan x(\tan x-1)(\tan x+1)=\tan x\cdot\dfrac{\sin^2x-\cos^2x}{\cos^2x}$. Expand the cosine: $\cos\left(x+\frac\pi4\right)=\dfrac{\cos x-\sin x}{\sqrt2}$. Since $\sin^2x-\cos^2x=-(\cos x-\sin x)(\cos x+\sin x)$, the factor $\cos x-\sin x$ cancels:

$$\frac{-\sqrt2\tan x(\cos x+\sin x)}{\cos^2x}\ \to\ \frac{-\sqrt2\cdot1\cdot\sqrt2}{\frac12}=-4.$$

Question 50

$\displaystyle\lim_{x\to\pi}\frac{1-\sin\frac x2}{\cos\frac x2\left(\cos\frac x4-\sin\frac x4\right)}$

Solution. Put $t=\pi-x$, so $t\to0$. Then $\frac x2=\frac\pi2-\frac t2$, which gives $\sin\frac x2=\cos\frac t2$ and $\cos\frac x2=\sin\frac t2$. Also $\frac x4=\frac\pi4-\frac t4$, and $\cos\left(\frac\pi4-u\right)-\sin\left(\frac\pi4-u\right)=\sqrt2\sin u$ with $u=\frac t4$.

Numerator: $1-\cos\frac t2=2\sin^2\frac t4$. So the whole expression is

$$\frac{2\sin^2\frac t4}{\sin\frac t2\cdot\sqrt2\sin\frac t4}=\frac{2\sin\frac t4}{\sqrt2\cdot2\sin\frac t4\cos\frac t4}=\frac{1}{\sqrt2\cos\frac t4}\ \to\ \frac1{\sqrt2}.$$

Question 51

Show that $\displaystyle\lim_{x\to4}\frac{|x-4|}{x-4}$ does not exist.

Solution. The modulus changes meaning on either side of $4$, so find the one-sided limits.

• For $x>4$ (right of $4$): $|x-4|=x-4$, so the quotient is $1$ and $\displaystyle\lim_{x\to4^+}=1$.
• For $x<4$ (left of $4$): $|x-4|=-(x-4)$, so the quotient is $-1$ and $\displaystyle\lim_{x\to4^-}=-1$.

Since $1\ne-1$, the limit does not exist.

Question 52

Let $f(x)=\begin{cases}\dfrac{k\cos x}{\pi-2x}, & x\ne\dfrac\pi2\\[2mm] 3, & x=\dfrac\pi2\end{cases}$ and $\displaystyle\lim_{x\to\frac\pi2}f(x)=f\left(\frac\pi2\right)$. Find $k$.

Solution. The condition says $\displaystyle\lim_{x\to\frac\pi2}\frac{k\cos x}{\pi-2x}=3$. Put $t=\frac\pi2-x\to0$. Then $\cos x=\sin t$ and $\pi-2x=2t$:

$$\lim_{t\to0}\frac{k\sin t}{2t}=\frac k2.$$

So $\dfrac k2=3$, giving $k=6$.

Question 53

Let $f(x)=\begin{cases}x+2, & x\le-1\\ cx^2, & x>-1\end{cases}$. Find $c$ if $\displaystyle\lim_{x\to-1}f(x)$ exists.

Solution. The book prints the first condition as $x\le1$, which is a misprint for $x\le-1$. As printed the two conditions overlap on $(-1,1]$ and the function would not be well defined; the question asks about the limit at $-1$, so the pieces must meet at $-1$. The key’s $c=1$ is consistent with the corrected version.

The limit exists when the left-hand and right-hand limits agree.

• Left: $\displaystyle\lim_{x\to-1^-}(x+2)=1$.
• Right: $\displaystyle\lim_{x\to-1^+}cx^2=c$.

Equating, $c=1$.

Choose the correct answer out of the 4 options given against each Exercise 54 to 76 (M.C.Q.). This page covers 54 to 64.

Question 54

$\displaystyle\lim_{x\to\pi}\dfrac{\sin x}{x-\pi}$ is

Solution. Put $t=x-\pi\to0$. Then $\sin x=\sin(\pi+t)=-\sin t$, so the limit is $-\dfrac{\sin t}{t}\to-1$.

Question 55

$\displaystyle\lim_{x\to0}\dfrac{x^2\cos x}{1-\cos x}$ is

Solution. Use $\dfrac{1-\cos x}{x^2}\to\dfrac12$. Then the expression is $\cos x\cdot\dfrac{x^2}{1-\cos x}\to1\cdot2=2$.

Question 56

$\displaystyle\lim_{x\to0}\dfrac{(1+x)^n-1}{x}$ is

Solution. With $y=1+x\to1$ the expression is $\dfrac{y^n-1^n}{y-1}\to n\cdot1^{n-1}=n$. (It is also the derivative of $(1+x)^n$ at $0$.)

Question 57

$\displaystyle\lim_{x\to1}\dfrac{x^m-1}{x^n-1}$ is

Solution. Divide top and bottom by $x-1$. The top tends to $m\cdot1^{m-1}=m$ and the bottom to $n\cdot1^{n-1}=n$, so the limit is $\dfrac mn$.

Question 58

$\displaystyle\lim_{x\to0}\dfrac{1-\cos4\theta}{1-\cos6\theta}$ is

Solution. The book writes $x\to0$ but the variable is $\theta$; the intended limit is $\theta\to0$. Using $1-\cos A=2\sin^2\frac A2$:

$$\frac{\sin^22\theta}{\sin^23\theta}=\left(\frac{\sin2\theta}{2\theta}\right)^2\left(\frac{3\theta}{\sin3\theta}\right)^2\cdot\frac{4}{9}\to\frac49.$$

Question 59

$\displaystyle\lim_{x\to0}\dfrac{\operatorname{cosec}x-\cot x}{x}$ is

Solution. Combine: $\operatorname{cosec}x-\cot x=\dfrac{1-\cos x}{\sin x}$. So the expression is $\dfrac{1-\cos x}{x^2}\cdot\dfrac{x}{\sin x}\to\dfrac12\cdot1=\dfrac12$. (Options (B) and (D) are both printed as $1$ in the book; neither is correct.)

Question 60

$\displaystyle\lim_{x\to0}\dfrac{\sin x}{\sqrt{x+1}-\sqrt{1-x}}$ is

Solution. Rationalise the denominator: $(x+1)-(1-x)=2x$. So the expression is $\dfrac{\sin x}{x}\cdot\dfrac{\sqrt{x+1}+\sqrt{1-x}}2\to1\cdot\dfrac{1+1}{2}=1$.

Question 61

$\displaystyle\lim_{x\to\frac\pi4}\dfrac{\sec^2x-2}{\tan x-1}$ is

(A) $3$ (B) $1$ (C) $0$ (D) $\sqrt2$

Solution. None of the four options is correct, and the key’s (D) is wrong. Use $\sec^2x=1+\tan^2x$, so the numerator is $\tan^2x-1=(\tan x-1)(\tan x+1)$ and the factor $\tan x-1$ cancels:

$$\lim_{x\to\frac\pi4}(\tan x+1)=1+1=2.$$

The value $2$ is not among the options. Since no option is right, none is marked above.

Question 62

$\displaystyle\lim_{x\to1}\dfrac{(\sqrt x-1)(2x-3)}{2x^2+x-3}$ is

Solution. Factorise $2x^2+x-3=(2x+3)(x-1)$ and write $\sqrt x-1=\dfrac{x-1}{\sqrt x+1}$. The factor $x-1$ cancels:

$$\lim_{x\to1}\frac{2x-3}{(\sqrt x+1)(2x+3)}=\frac{-1}{2\cdot5}=-\frac1{10}.$$

Question 63

If $f(x)=\begin{cases}\dfrac{\sin[x]}{[x]}, & [x]\ne0\\ 0, & [x]=0\end{cases}$, where $[\,\cdot\,]$ denotes the greatest integer function, then $\displaystyle\lim_{x\to0}f(x)$ is equal to

Solution. Look at the two sides of $0$ separately.

• For $0<x<1$ we have $[x]=0$, so $f(x)=0$. The right-hand limit is $0$.
• For $-1<x<0$ we have $[x]=-1\ne0$, so $f(x)=\dfrac{\sin(-1)}{-1}=\sin1\approx0.84$. The left-hand limit is $\sin1$.

Since $0\ne\sin1$, the limit does not exist, so none of (A), (B), (C) can be right.

Question 64

$\displaystyle\lim_{x\to0}\dfrac{|\sin x|}{x}$ is

Solution. For small $x>0$, $\sin x>0$ and the quotient is $\dfrac{\sin x}{x}\to1$. For small $x<0$, $\sin x<0$ so $|\sin x|=-\sin x$ and the quotient is $-\dfrac{\sin x}{x}\to-1$. The two one-sided limits differ, so the limit does not exist.

Common mistakes

• Question 51 and 64. Treating a modulus quotient as the same as the quotient without the modulus. $\frac{|x-4|}{x-4}$ is $+1$ on one side and $-1$ on the other; students who reason “the modulus cancels” report $1$.
• Question 52. Substituting $x=\frac\pi2$ in $\frac{k\cos x}{\pi-2x}$ and stopping at $\frac00$. The value $f\left(\frac\pi2\right)=3$ is a separate piece; the limit is found from the other piece.
• Question 53. Equating the two pieces at $x=-1$ as if $f(-1)=c$, using the wrong piece for the left-hand limit. The left-hand limit uses $x+2$, which is $1$.
• Question 61. Choosing an option because it looks close to a plausible value. The limit is $2$; when no option fits, say so.
• Question 62. Cancelling $x-1$ without first rationalising $\sqrt x-1$; the factor is hidden inside the root.
• Question 63. Taking $\lim_{x\to0}\frac{\sin[x]}{[x]}$ to be $1$ by the standard limit. $[x]$ is $0$ or $-1$ near $0$ and never tends to zero in the sense the standard limit needs; the one-sided limits are different.

Practise next

• Other pages in this unit: Short Answer Type Q1–14, Short Answer Type Q15–28, Short and Long Answer Type Q29–46 and Objective and Fill in the Blanks Q65–80.
• NCERT textbook: Exercise 12.1, Exercise 12.2 and the Miscellaneous Exercise.