## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to0^+}\tan^{-1}(\ln x)=\frac{-\pi}{2}$$
$$A=\lim\limits_{x\to0^+}\tan^{-1}(\ln x)$$$$A=\lim\limits_{x\to0^+}\arctan(\ln x)$$ Let $t=\ln x$ As $x\to{0^+}$, $\ln x$ approaches $-\infty$. Therefore $t\to{-\infty}$ $$A=\lim\limits_{t\to-\infty}\arctan t$$$$A=\frac{-\pi}{2}$$ *NOTES TO REMEMBER: $\lim\limits_{x\to-\infty}\arctan x=\frac{-\pi}{2}$