JEE Advanced 2018 Paper 1 · Mathematics · Limits · hard

Evaluate L=lim x→0((tan x)/(x)·(sin x)/(x))^1/x^2.

Numerical answer

Evaluate

L=\displaystyle\lim_{x\to0}\left(\frac{\tan x}{x}\cdot\frac{\sin x}{x}\right)^{1/x^2}

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 1.18

The expression has the form

1^\infty

, so take logarithms:

\ln L=\lim_{x\to0}\frac{\ln\frac{\tan x}x+\ln\frac{\sin x}x}{x^2}

. Use the expansions

\frac{\tan x}{x}=1+\frac{x^2}3+O(x^4)

and

\frac{\sin x}x=1-\frac{x^2}6+O(x^4)

. Then

\ln\frac{\tan x}{x}=\frac{x^2}3+O(x^4)

and

\ln\frac{\sin x}{x}=-\frac{x^2}6+O(x^4)

. The numerator is

\frac{x^2}{3}-\frac{x^2}6=\frac{x^2}{6}

to leading order, so

\ln L=\frac16

and

L=e^{1/6}\approx1.1814

. Numerical evaluation at

x=10^{-3}

gives

1.18136

, matching. The answer is

1.18

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