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 : $1$

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.

• @sultan : I would also recommend substituting $s = 1/x$ and letting $s \to 0^+$. This will make the derivatives much simpler. I encourage students to do this but they rarely do. Apr 29, 2013 at 15:32

$$\displaystyle\lim_{x\to\infty }\left(\tan^{-1}\left(\frac{1}{x}\right) \right )^{\frac{1}{x}}=\lim_{x\to\infty }e^{\frac{\ln\left(\tan^{-1}\left(\frac{1}{x}\right)\right)}{x}}=\Bigg[\tan^{-1}\left(\frac{1}{x}\right)=u\Bigg/\tan\implies \frac{1}{x}=\tan u\Bigg]$$ $$x\to\infty\implies u\to 0^+$$

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 $$\lim_{x\to\infty }\left(\tan^{-1}\left(\frac{1}{x}\right)\right)^{\frac{1}{x}}=1$$

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.