Question

$\displaystyle\lim _{x\rightarrow \infty }\left( \dfrac {\pi } {2}-\arctan x\right) ^{\dfrac {1} {\ln x}}=?$

Answer 1

$$\lim_{x\rightarrow\infty}\left(\frac{\pi}{2}-arctanx\right)^{\frac{1}{\ln x}}=\lim_{x\rightarrow\infty}e^{\ln \left(\frac{\pi}{2}-arctanx\right)^{\frac{1}{\ln x}}}=\lim_{x\rightarrow\infty}e^{{\frac{1}{\ln x}}\ln \left(\frac{\pi}{2}-arctanx\right)}$$

$$=$$

$$\lim_{x\rightarrow\infty}e^{{\frac{\ln \left(\frac{\pi}{2}-arctanx\right)}{\ln x}}}=$$

$$=$$

$$e^{\lim_{x\rightarrow\infty}{{\frac{\ln \left(\frac{\pi}{2}-arctanx\right)}{\ln x}}}}$$

$$=$$

$$\ldots$$

L'Hospital uygulayarak sonucu bulursun.