NCERT Exemplar Solutions Class 11 Maths Limits and Derivatives — Short Answer Type Q15–28

NCERT Exemplar Problems, Class 11 Mathematics — Limits and Derivatives, Short Answer Type questions 15 to 28. All 14 questions solved.

Fourteen trigonometric limits. They all run on three standard limits: $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$, $\displaystyle\lim_{x\to0}\frac{\tan x}{x}=1$ and $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$, plus the identities that bring an expression into those shapes ($1-\cos2x=2\sin^2x$, $\sin x-\cos x=\sqrt2\sin(x-\frac\pi4)$, and so on). When the limit point is not $0$, shift the variable so that it is.

One question is misprinted (Question 15) and one printed key answer is wrong (Question 20). Both are explained where they occur.

Key insight. Every limit here is a disguised $\dfrac{\sin\theta}{\theta}\to1$ with $\theta\to0$. Rewrite the expression so that the quantity inside each sine or cosine is also the quantity in the denominator; if the limit point is not $0$, substitute $t=x-\text{limit point}$ first.

Evaluate:

Question 15

$\displaystyle\lim_{x\to a}\frac{\sin3x}{\sin7x}$

Solution. As printed, $x\to a$, not $x\to0$, and the key’s $\frac37$ is the value for $x\to0$. For a general constant $a$ with $\sin7a\ne0$ both $\sin3x$ and $\sin7x$ are continuous at $a$, so the limit is just the value:

$$\lim_{x\to a}\frac{\sin3x}{\sin7x}=\frac{\sin3a}{\sin7a}.$$

If $x\to0$ was meant, divide top and bottom by $x$: $\dfrac{3\cdot\frac{\sin3x}{3x}}{7\cdot\frac{\sin7x}{7x}}\to\dfrac37$.

Question 16

$\displaystyle\lim_{x\to0}\frac{\sin^22x}{\sin^24x}$

Solution. Write each sine against its own argument:

$$\frac{\sin^22x}{\sin^24x}=\left(\frac{\sin2x}{2x}\right)^2\left(\frac{4x}{\sin4x}\right)^2\cdot\frac{(2x)^2}{(4x)^2}\to1\cdot1\cdot\frac14=\frac14.$$

Question 17

$\displaystyle\lim_{x\to0}\frac{1-\cos2x}{x^2}$

Solution. $1-\cos2x=2\sin^2x$, so the expression is $2\left(\dfrac{\sin x}{x}\right)^2\to2$.

Question 18

$\displaystyle\lim_{x\to0}\frac{2\sin x-\sin2x}{x^3}$

Solution. Use $\sin2x=2\sin x\cos x$, which gives $2\sin x-\sin2x=2\sin x(1-\cos x)$. Then

$$\frac{2\sin x(1-\cos x)}{x^3}=2\cdot\frac{\sin x}{x}\cdot\frac{1-\cos x}{x^2}\to2\cdot1\cdot\frac12=1.$$

Question 19

$\displaystyle\lim_{x\to0}\frac{1-\cos mx}{1-\cos nx}$

Solution. Write $1-\cos\theta=2\sin^2\frac\theta2$ in both places:

$$\frac{\sin^2\frac{mx}2}{\sin^2\frac{nx}2}=\left(\frac{\sin\frac{mx}2}{\frac{mx}2}\right)^2\left(\frac{\frac{nx}2}{\sin\frac{nx}2}\right)^2\cdot\frac{m^2}{n^2}\to\frac{m^2}{n^2}.$$

Question 20

$\displaystyle\lim_{x\to\frac\pi3}\frac{\sqrt{1-\cos6x}}{\sqrt2\left(\frac\pi3-x\right)}$

Solution. The printed key says $3$, but the limit does not exist. The root hides a modulus: $1-\cos6x=2\sin^23x$, so $\sqrt{1-\cos6x}=\sqrt2\,|\sin3x|$, not $\sqrt2\sin3x$. The expression is then $\dfrac{|\sin3x|}{\frac\pi3-x}$.

Put $t=\frac\pi3-x$, so $x\to\frac\pi3$ means $t\to0$ and $\sin3x=\sin(\pi-3t)=\sin3t$. The expression is $\dfrac{|\sin3t|}{t}$.

• As $t\to0^+$ (that is, $x\to\frac\pi3$ from the left): $\dfrac{\sin3t}{t}\to3$.
• As $t\to0^-$ (that is, $x\to\frac\pi3$ from the right): $\sin3t<0$, so $|\sin3t|=-\sin3t$ and the expression tends to $-3$.

The one-sided limits are $3$ and $-3$, so the two-sided limit does not exist. The key’s $3$ is the left-hand limit, which is what you get if the modulus is dropped by mistake.

Question 21

$\displaystyle\lim_{x\to\frac\pi4}\frac{\sin x-\cos x}{x-\frac\pi4}$

Solution. Combine the numerator into one sine: $\sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right)$. With $t=x-\frac\pi4\to0$ the expression is $\sqrt2\,\dfrac{\sin t}{t}\to\sqrt2$.

Question 22

$\displaystyle\lim_{x\to\frac\pi6}\frac{\sqrt3\sin x-\cos x}{x-\frac\pi6}$

Solution. The coefficients $\sqrt3$ and $1$ are $2\cos\frac\pi6$ and $2\sin\frac\pi6$, so $\sqrt3\sin x-\cos x=2\left(\sin x\cos\frac\pi6-\cos x\sin\frac\pi6\right)=2\sin\left(x-\frac\pi6\right)$. Then, with $t=x-\frac\pi6$, the limit is $2\cdot\dfrac{\sin t}{t}\to2$.

Question 23

$\displaystyle\lim_{x\to0}\frac{\sin2x+3x}{2x+\tan3x}$

Solution. Divide numerator and denominator by $x$:

$$\frac{2\cdot\frac{\sin2x}{2x}+3}{2+3\cdot\frac{\tan3x}{3x}}\to\frac{2+3}{2+3}=1.$$

Question 24

$\displaystyle\lim_{x\to a}\frac{\sin x-\sin a}{\sqrt x-\sqrt a}$

Solution. Multiply and divide by $\sqrt x+\sqrt a$ so that the denominator becomes $x-a$:

$$\frac{\sin x-\sin a}{x-a}\cdot\left(\sqrt x+\sqrt a\right).$$

The first factor is the derivative of $\sin$ at $a$, namely $\cos a$ (or use $\sin x-\sin a=2\cos\frac{x+a}2\sin\frac{x-a}2$). So the limit is $2\sqrt a\cos a$.

Question 25

$\displaystyle\lim_{x\to\frac\pi6}\frac{\cot^2x-3}{\operatorname{cosec}x-2}$

Solution. Use $\cot^2x=\operatorname{cosec}^2x-1$ so that the numerator becomes $\operatorname{cosec}^2x-4=(\operatorname{cosec}x-2)(\operatorname{cosec}x+2)$. The factor that vanishes at $x=\frac\pi6$ (where $\operatorname{cosec}x=2$) cancels:

$$\lim_{x\to\frac\pi6}\left(\operatorname{cosec}x+2\right)=2+2=4.$$

Question 26

$\displaystyle\lim_{x\to0}\frac{\sqrt2-\sqrt{1+\cos x}}{\sin^2x}$

Solution. Rationalise the numerator: $2-(1+\cos x)=1-\cos x$. Also $\sin^2x=(1-\cos x)(1+\cos x)$, so the factor $1-\cos x$ cancels:

$$\frac{1-\cos x}{(1-\cos x)(1+\cos x)\left(\sqrt2+\sqrt{1+\cos x}\right)}=\frac{1}{(1+\cos x)\left(\sqrt2+\sqrt{1+\cos x}\right)}\to\frac{1}{2\cdot2\sqrt2}=\frac1{4\sqrt2}.$$

Question 27

$\displaystyle\lim_{x\to0}\frac{\sin x-2\sin3x+\sin5x}{x}$

Solution. Each term divided by $x$ tends to its coefficient of $x$ inside the sine: $\dfrac{\sin x}{x}\to1$, $\dfrac{2\sin3x}{x}\to6$, $\dfrac{\sin5x}{x}\to5$. The total is $1-6+5=0$. (Alternatively, $\sin x+\sin5x=2\sin3x\cos2x$, so the numerator is $-2\sin3x(1-\cos2x)$, which is of order $x^3$, so the quotient tends to $0$.)

Question 28

If $\displaystyle\lim_{x\to1}\frac{x^4-1}{x-1}=\lim_{x\to k}\frac{x^3-k^3}{x^2-k^2}$, find $k$.

Solution. Use $\dfrac{x^n-a^n}{x-a}\to na^{n-1}$. The left side is $4\cdot1^3=4$. For the right side divide top and bottom by $x-k$:

$$\lim_{x\to k}\frac{\frac{x^3-k^3}{x-k}}{\frac{x^2-k^2}{x-k}}=\frac{3k^2}{2k}=\frac{3k}2\quad(k\ne0).$$

Equate: $\dfrac{3k}2=4$, so $k=\dfrac83$. (The value $k=0$ gives a right-hand limit of $0$, not $4$, so it is rejected.)

Common mistakes

• Question 20. Writing $\sqrt{1-\cos6x}=\sqrt2\sin3x$ and forgetting the modulus. A square root is always non-negative, so it equals $|\sin3x|$; dropping that is exactly how the printed key reaches $3$.
• Question 15. Applying $\frac{\sin\theta}{\theta}\to1$ when the limit is not $0$. That standard limit needs the angle to tend to $0$; at $x\to a$ you just substitute.
• Question 16. Cancelling “$\sin$” or the squares directly: $\frac{\sin^22x}{\sin^24x}$ is not $\frac{2x}{4x}$ squared until each sine has been matched with its own argument.
• Question 21 and 22. Forgetting to change variable. At $x\to\frac\pi4$ the standard limit does not apply until the numerator is turned into a single $\sin(x-\frac\pi4)$.
• Question 25. Substituting $x=\frac\pi6$ and stopping at $\frac00$. The key step is recognising $\cot^2x=\operatorname{cosec}^2x-1$.
• Question 28. Taking $\lim_{x\to k}\frac{x^3-k^3}{x^2-k^2}$ as $\frac{k^3-k^3}{k^2-k^2}=\frac00$ and giving up; it is $\frac{3k}{2}$, and it is only valid for $k\neq0$.

Practise next

• Other pages in this unit: Short Answer Type Q1–14, Short and Long Answer Type Q29–46, Long Answer and Objective Type Q47–64 and Objective and Fill in the Blanks Q65–80.
• NCERT textbook: Exercise 12.1, Exercise 12.2 and the Miscellaneous Exercise.