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Limits Continuity and Differentiability question

2024 · 31 Jan · Shift 1 · Q31

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Limits Continuity and Differentiability
2024 · 31 Jan · Shift 1 · Q31

Limits Continuity and Differentiability question

2024 · 31 Jan · Shift 1 · Q31

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

\lim_{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}

A
is equal to 1
B
does not exist
C
is equal to $-1$
D
is equal to 2

View written solutionFree

Correct answer: D

We need to evaluate

L=\lim_{x\to 0}\frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}.

Let $t=|\sin x|.$ As $x\to 0$ , we have $t\to 0$ . Then the numerator becomes $e^{2t}-2t-1.$
Use the expansion of $e^{2t}$ about $t=0$ :

e^{2t}=1+2t+\frac{(2t)^2}{2!}+\frac{(2t)^3}{3!}+\cdots =1+2t+2t^2+O(t^3).

Therefore,

e^{2t}-2t-1=2t^2+O(t^3).

So,

\frac{e^{2|\sin x|}-2|\sin x|-1}{x^2} =\frac{2|\sin x|^2+O(|\sin x|^3)}{x^2}.

Since $|\sin x|^2=\sin^2 x$ ,

=2\frac{\sin^2 x}{x^2}+\frac{O(|\sin x|^3)}{x^2}.

Now as $x\to 0$ , $\frac{\sin x}{x}\to 1,$ and since $|\sin x|\sim |x|$ , we get

L=2\lim_{x\to 0}\frac{\sin^2 x}{x^2}=2\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2=2.

Hence,

L=2\lim_{x\to 0}\frac{\sin^2 x}{x^2}=2\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2=2.

Therefore the limit exists and is equal to $2$ .
Checking options:

A: $1$ — incorrect
B: does not exist — incorrect
C: $-1$ — incorrect
D: $2$ — correct

Hence, the correct option is D.

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