Answer
$$\lim\limits_{x\to0^+}\tan^{-1}(1/x)=\frac{\pi}{2}$$
Work Step by Step
$$A=\lim\limits_{x\to0^+}\tan^{-1}(1/x)$$
Let $u=1/x$
As $x\to0^+$, $1/x$ approaches $\infty$. That means $u$ approaches $\infty$, too.
Therefore, $$A=\lim\limits_{u\to\infty}\tan^{-1}u=\frac{\pi}{2}$$
*NOTES TO REMEMBER:
$$\lim\limits_{x\to\infty}\tan^{-1}x=\frac{\pi}{2}$$ and $$\lim\limits_{x\to-\infty}\tan^{-1}x=-\frac{\pi}{2}$$
