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  • Find the limit of the following: <img alt="\lim_{x\rightarrow 0}\left [ \frac{1}{1+x} \right ]^{-\cot \left ( \frac{\pi x}{2} \right )}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim_

Find the limit of the following: \lim_{x\rightarrow 0}\left [ \frac{1}{1+x} \right ]^{-\cot \left ( \frac{\pi x}{2} \right )}

Option: 1

1


Option: 2

0


Option: 3

e^{-\frac{2}{\pi}}


Option: 4

e


Answers (1)

best_answer

Given: \lim_{x\rightarrow 0}\left [ \frac{1}{1+x} \right ]^{-\cot \left ( \frac{\pi x}{2} \right )}

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 _{x \rightarrow a}(f(x)-1)(g(x))}.

=\lim _{x \rightarrow 0}\left[1+\left(\frac{1}{1+x}-1\right)\right]^{-\cot (\pi x / 2)}
=e^{\lim _{x \rightarrow 0}\left[\left(\frac{1}{1+x}-1\right)\right]-\cot (\pi x / 2)}

Applying trigonometric formula \cot x= \frac{1}{\tan x}, then the equation becomes

= e^{\lim_{x\rightarrow 0}\left [ \frac{x}{1+x} \right ]\frac{1}{\tan \left ( \frac{\pi x}{2} \right )}}

Multiply and divide by \frac{\pi}{2}.

= e^{\lim_{x\rightarrow 0}\left [ \frac{-1}{1+x} \right ]\frac{\pi x/2}{\tan \left ( \frac{\pi x}{2} \right )}\frac{2}{\pi}}
=e^{-\frac{2}{\pi}}
Therefore, \lim_{x\rightarrow 0}\left [ \frac{1}{1+x} \right ]^{-\cot \left ( \frac{\pi x}{2} \right )}= e^{-\frac{2}{\pi}}.


 

Posted by

Ritika Harsh

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