## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to\infty}\frac{\ln x}{\sqrt x}=0$$
$$A=\lim_{x\to\infty}\frac{\ln x}{\sqrt x}$$ As $x\to\infty$, $\ln x$ approaches $\infty$ and $\sqrt x$ also approaches $\infty$. So this limit is in an indeterminate form of $\infty/\infty$. We can use L'Hospital's Rule here: $$A=\lim_{x\to\infty}\frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(\sqrt x)}$$ $$A=\lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{2\sqrt x}}$$ $$A=\lim_{x\to\infty}\frac{2\sqrt x}{x}$$ Divide both numerator and denominator by $x$, we have $$A=\lim_{x\to\infty}\frac{\frac{2\sqrt x}{x}}{\frac{x}{x}}$$ $$A=\lim_{x\to\infty}\frac{2\sqrt{\frac{x}{x^2}}}{1}$$ $$A=\lim_{x\to\infty}2\sqrt{\frac{1}{x}}$$ We remember that $\lim_{x\to\infty}(\frac{1}{x})=0$. So, $$A=2\sqrt0$$ $$A=0$$