JEE Advanced 2019 Paper 2 · Mathematics · Limits · medium

The value of lim x→0((sin x+tan x)/(2x))^1/x^2 is ______.

Numerical answer

The value of

\displaystyle\lim_{x\to0}\left(\dfrac{\sin x+\tan x}{2x}\right)^{1/x^{2}}

is ______. (Give the answer correct to two decimal places.)

Show answer and solution

Answer: 1.09

The base tends to

1

, giving the

1^\infty

form, so the limit is

e^{L}

with

L=\lim_{x\to0}\frac{1}{x^2}\left(\frac{\sin x+\tan x}{2x}-1\right)

. Use the expansions

\frac{\sin x}{x}=1-\frac{x^2}6+\cdots

and

\frac{\tan x}{x}=1+\frac{x^2}3+\cdots

. Then

\frac{\sin x+\tan x}{2x}=1+\frac12\left(-\frac{x^2}6+\frac{x^2}3\right)+O(x^4)=1+\frac{x^2}{12}+O(x^4)

. Hence

L=\frac1{12}

and the limit is

e^{1/12}\approx1.0869

, i.e.

1.09

. Evaluating the expression numerically at

x=0.01

gives

1.08691

, consistent with this result.

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