Find the value of $\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}$ Can this limit be solved without using L'Hopital's rule : 
$$\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}$$
Answer of this limit is : $1$
 A: Hint: $$\frac{\pi}2-\tan^{-1}(x)=\tan^{-1}\left(\frac{1}x\right)$$
So $$\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\large\frac{1}{x}} = \lim_{x\to \infty} \left(\tan^{-1}\left(\frac{1}{x}\right)\right)^{1/x}$$
If you're familiar with L'hopital, I'd recommend it.
A: $\displaystyle\lim_{x\to\infty }\left(\tan^{-1}\left(\frac{1}{x}\right) \right )^{\frac{1}{x}}=\lim_{x\to\infty }e^{\frac{\ln\left(\tan^{-1}\left(\frac{1}{x}\right)\right)}{x}}=\Bigg[\tan^{-1}\left(\frac{1}{x}\right)=u\Bigg/\tan\implies \frac{1}{x}=\tan u\Bigg]$
$$x\to\infty\implies u\to 0^+$$ 
then $$\lim_{x\to\infty }\left(\tan^{-1}\left(\frac{1}{x}\right)\right)^{\frac{1}{x}}=\lim_{u\to 0^+}u^{\tan (u)}=\lim_{u\to 0^+}e^{\tan(u)\ln(u)}$$
$$\lim_{u\to 0^+}e^{\tan(u) \ln(u)}=\lim_{u\to 0^+}e^{\frac{\tan(u)}{u} u\ln(u)}$$
we know that $$\lim_{u\to 0^+}\frac{\tan(u)}{u}=1\;\&\;\lim_{u\to 0^+}u\ln(u)=0$$
so $$\lim_{u\to 0^+}e^{\frac{\tan(u)}{u} u\ln(u)}=e^{0}=1$$
so $$\lim_{x\to\infty }\left(\tan^{-1}\left(\frac{1}{x}\right)\right)^{\frac{1}{x}}=1$$
A: Here is another approach
$$ \lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}=\lim_{x \rightarrow \infty}e^{\frac{\ln\left(\frac{\pi}{2}-\tan^{-1}x\right)}{x}} =e^{\lim_{n\to \infty}\frac{\ln\left(\frac{\pi}{2}-\tan^{-1}x\right)}{x}} $$
$$ = e^{\lim_{n\to \infty}\frac{ -\frac{1}{1+x^2} }{1}}=e^0=1.$$
Notice that, We used the L'hopital's rule.
