Question: Limit of $\lim\limits_{n \to \infty} (1+ \frac{\pi}{2} – \arctan(x))^x$
The first things I notice are : $\lim\limits_{x \to \infty} \arctan(x) = \frac{\pi}{2}$ and the limit looks something of the form $(1 + \frac{1}{x})^x$. Unfortunetly, I can not seem to apply these ideas to solve the limits. I am not sure if it is right:
$$\ln(y) = \lim\limits_{x \to \infty} x \ln(1 + \frac{\pi}{2} – \arctan(x)) = \lim\limits_{x \to \infty}\frac{\ln(1+\frac{\pi}{2}- \arctan(x))}{\frac{1}{x}}$$
Apply l'hopital's rule … ?
Could someone confirm this approach is right, or if wrong provide a correct approach?
Best Answer
Since $\frac{\pi}{2} - \arctan(x) =\arctan \left(\frac1x\right)\to 0$ we can use that
$$\left(1+ \frac{\pi}{2} - \arctan(x)\right)^x=\left[\left(1+ \arctan\left(\frac1x\right)\right)^{\frac{1}{\arctan\left(\frac1x\right)}}\right]^{x\arctan\left(\frac1x\right)}$$
and then refer to standard limits.
Or as an alternative, following your idea
$$\lim\limits_{x \to \infty}\frac{\ln(1+\frac{\pi}{2}- \arctan(x))}{\frac{1}{x}}=\lim\limits_{x \to \infty}\frac{\ln\left(1+\arctan \left(\frac1x\right)\right)}{\arctan \left(\frac1x\right)}\,\frac{\arctan \left(\frac1x\right)}{\frac1x}$$
and then conclude again by standard limits.