## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to0^+}\tan^{-1}(1/x)=\frac{\pi}{2}$$
$$A=\lim\limits_{x\to0^+}\tan^{-1}(1/x)$$ 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 $$\lim\limits_{x\to-\infty}\tan^{-1}x=-\frac{\pi}{2}$$