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  • Evaluate <img alt="\mathrm{ \lim _{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan ^{-1} x\right)^{\frac{1}{x}} }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Clim

Evaluate

\mathrm{ \lim _{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan ^{-1} x\right)^{\frac{1}{x}} }

Option: 1

1 / 2


Option: 2

0


Option: 3

2


Option: 4

1


Answers (1)

best_answer

\mathrm{\lim _{x \rightarrow \infty}\left(\tan ^{-1}\left(\frac{1}{x}\right)\right)^{\frac{1}{z}}=\lim _{u \rightarrow 0^{+}} u^{\tan (u)}=\lim _{u \rightarrow 0^{+}} e^{\tan (u) \ln (u)}}

                           \mathrm{\lim _{u \rightarrow 0^{+}} e^{\tan (u) \ln (u)}=\lim _{u \rightarrow 0^{+}} e^{\frac{\tan (u)}{u} u \ln (u)}}

we know that

                          \mathrm{\lim _{u \rightarrow 0^{+}} \frac{\tan (u)}{u}=1 \& \lim _{u \rightarrow 0^{+}} u \ln (u)=0}

So                           

                             \mathrm{\lim _{u \rightarrow 0^{+}} e^{\frac{\tan (u)}{u} u \ln (u)}=e^0=1}

So         

                     \mathrm{\lim _{x \rightarrow \infty}\left(\tan ^{-1}\left(\frac{1}{x}\right)\right)^{\frac{1}{x}}=1}

Posted by

Nehul

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