## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to\infty}\frac{\ln\sqrt x}{x^2}=0$$
$$A=\lim_{x\to\infty}\frac{\ln\sqrt x}{x^2}$$ As $x\to\infty$, $\ln\sqrt x\to\infty$ and $x^2\to\infty$. So this is an indeterminate form of $\infty/\infty$. So L'Hospital's Rule can be applied. $$A=\lim_{x\to\infty}\frac{(\ln\sqrt x)'}{(x^2)'}$$ $$A=\lim_{x\to\infty}\frac{\frac{1}{\sqrt x}(\sqrt x)'}{2x}$$ $$A=\lim_{x\to\infty}\frac{\frac{1}{\sqrt x}\frac{1}{2\sqrt x}}{2x}$$ $$A=\lim_{x\to\infty}\frac{\frac{1}{2x}}{2x}$$ $$A=\lim_{x\to\infty}\frac{1}{4x^2}$$ $$A=\frac{1}{4}\lim_{x\to\infty}\frac{1}{x^2}$$ $$A=\frac{1}{4}\times0=0$$ (since $\lim_{x\to\infty}\frac{1}{x^2}=0$)