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  • Evaluate the limit of the following: <img alt="\lim _{x \rightarrow \frac{\pi}{2}-0}\left[1+\left(1-\sin ^3 x\right)\right]^{1 / 3 \cos ^2 x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%

Evaluate the limit of the following: \lim _{x \rightarrow \frac{\pi}{2}-0}\left[1+\left(1-\sin ^3 x\right)\right]^{1 / 3 \cos ^2 x}

Option: 1

\sqrt{e}


Option: 2

e^{1/\sqrt{a}}


Option: 3

e^{a}


Option: 4

e


Answers (1)

best_answer

Given:\lim _{x \rightarrow \frac{\pi}{2}-0}\left[1+\left(1-\sin ^3 x\right)\right]^{1 / 3 \cos ^2 x}

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 \frac{\pi}{2}-0} \frac{1-\sin ^3 x}{3 \cos ^2 x}}

Apply the change of variable rule, the equation becomes

= e^{\lim _{h \rightarrow 0} \frac{1-\sin ^3\left(\frac{\pi}{2}-h\right)}{3 \cos ^2\left(\frac{\pi}{2}-h\right)}}
=e^{\lim _{h \rightarrow 0} \frac{1-\cos ^3 h}{3 \sin ^2 h}}
=e^{\lim _{h \rightarrow 0} \frac{(1-\cos h)\left(1+\cos h+\cos ^2 h\right)}{3(1-\cos h)(1+\cos h)}}
=e^{\lim _{h \rightarrow 0} \frac{1+\cos h+\cos ^2 h}{3(1+\cos h)}}
=e^{\frac{3}{3\left ( 2 \right )}}
=e^{\frac{1}{2}}
=\sqrt{e}

Therefore,\lim _{x \rightarrow \frac{\pi}{2}-0}\left[1+\left(1-\sin ^3 x\right)\right]^{1 / 3 \cos ^2 x}=\sqrt{e}.

 

Posted by

Devendra Khairwa

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