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  • Find the limit of the following:<img alt="\lim _{x \rightarrow 0}\left[\frac{1}{1-\tan ^2 x}\right]^{x \tan x / \sin ^4 x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5C

Find the limit of the following:\lim _{x \rightarrow 0}\left[\frac{1}{1-\tan ^2 x}\right]^{x \tan x / \sin ^4 x}

Option: 1

1


Option: 2

0


Option: 3

\infty


Option: 4

e


Answers (1)

best_answer

Given: \lim _{x \rightarrow 0}\left[\frac{1}{1-\tan ^2 x}\right]^{x \tan x / \sin ^4 x}

If \lim _{x \rightarrow a} f(x)=1$ and $\lim _{x \rightarrow a} g(x)=\infty$, then $\lim _{x \rightarrow a} f(x)^{g(x)}=e^{\lim ^{\lim (f(x)-1)(g(x))}}.
=\lim _{x \rightarrow 0}\left[1+\left(\frac{1}{1-\tan ^2 x}-1\right)\right]^{x \tan x / \sin ^4 x}
=e^{\lim _{x \rightarrow 0}\left(\frac{1}{1-\tan ^2 x}-1\right) \frac{x \tan x}{\sin ^4 x}}
=e^{\lim _{x \rightarrow 0}\left(\frac{\tan ^2 x}{1-\tan ^2 x}\right) \frac{x \tan x}{\sin ^4 x}}

Multiply and divide the numerator and denominator by x^3, the equation becomes,
=e^{\lim _{x \rightarrow 0}\left ( \frac{x^{4}\frac{\tan x}{x}\frac{\tan ^{2}x}{x^{2}}}{x^{4}\frac{\sin ^{4}x}{x^{4}}\left ( 1-\tan ^{2}x\right )} \right )}
=e^{\lim _{x \rightarrow 0}\frac{\frac{\tan x}{x}\frac{\tan ^{2}x}{x^{2}}}{\frac{\sin ^{4}x}{x^{4}}\left (1-\tan ^{2}x \right )}}

Using the properties of limit \lim _{x \rightarrow 0} \frac{\tan x}{x}=1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1,then the expression becomes,

= e^{\frac{1}{1}}
= e
Therefore,\lim _{x \rightarrow 0}\left[\frac{1}{1-\tan ^2 x}\right]^{x \tan x / \sin ^4 x}= e.

Posted by

shivangi.bhatnagar

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