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  • Find the limit of the following:  <img alt="\lim _{x \rightarrow \pi / 2}[1+(\cot x-\cos x)]^{1 /(\pi-2 x)^3}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarr

Find the limit of the following:  \lim _{x \rightarrow \pi / 2}[1+(\cot x-\cos x)]^{1 /(\pi-2 x)^3}

Option: 1

e^{\frac{1}{16}}


Option: 2

0


Option: 3

\frac{2}{n}


Option: 4

e


Answers (1)

best_answer

Given: \lim _{x \rightarrow \pi / 2}[1+(\cot x-\cos x)]^{1 /(\pi-2 x)^3}

If \lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0$, then $e^{\lim _{x \rightarrow a} \frac{f(x)}{g(x)}}.

=e^{\lim _{x \rightarrow \pi / 2} \frac{\cot x-\cos x}{(\pi-2 x)^3}}
=e^{\lim _{x \rightarrow \pi / 2} \frac{\cot x(1-\sin x)}{-8(x-\pi / 2)^3}}

Apply the trigonometric ratios \cot x=\tan \left(90^{\sharp}-x\right)$ and $\sin x=\cos \left(90^{\sharp}-x\right),the equation becomes,

=e^{\lim _{x \rightarrow \pi / 2} \frac{\tan (\pi / 2-x)\left[1-\cos \left(\pi / 2^{-x}\right)\right]}{8\left(\pi / 2^{-x}\right)(\pi / 2-x)^2}}

Again, use the trigonometric formula 1-\cos 2 x=2 \sin ^2 x, the equation becomes,

=e^{\lim _{x \rightarrow \pi/ 2} \frac{\tan \left(\pi / 2^{-x}\right)\left[\sin ^2(\pi / 2-x)\right]}{8(\pi / 2-x) 2(\pi / 2-x)^2}}
=e^{1 / 16}

Use the limit formula \lim _{x \rightarrow 0} \frac{\sin x}{x}=1.

= e^{\lim_{x\rightarrow \pi }2\cos \left ( x/2 \right )}
= e^{0}
= 1

Therefore, \lim _{x \rightarrow \pi / 2}[1+(\cot x-\cos x)]^{1 /(\pi-2 x)^3}= e^{1/16}.

Posted by

Pankaj Sanodiya

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