## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to\infty}\frac{(\ln x)^2}{x}=0$$
$$A=\lim_{x\to\infty}\frac{(\ln x)^2}{x}$$ As $x$ approaches $\infty$, $(\ln x)^2$ would approach $\infty$ and $x$ would also approach $\infty$. So we have an indeterminate form of $\infty/\infty$. To deal with this, we apply the L'Hospital's Rule as follows: $$A=\lim_{x\to\infty}\frac{[(\ln x)^2]'}{x'}$$ $$A=\lim_{x\to\infty}\frac{2\ln x(\ln x)'}{1}$$ $$A=\lim_{x\to\infty}\frac{2\ln x\frac{1}{x}}{1}$$ $$A=\lim_{x\to\infty}\frac{2\ln x}{x}$$ Again, as $x$ approaches $\infty$, $2\ln x$ would approach $\infty$ and $x$ would also approach $\infty$. We still have an indeterminate form of $\infty/\infty$ to use L'Hospital's Rule another time: $$A=\lim_{x\to\infty}\frac{(2\ln x)'}{x'}$$ $$A=\lim_{x\to\infty}\frac{\frac{2}{x}}{1}$$ $$A=\lim_{x\to\infty}\frac{2}{x}$$ As $x\to\infty$, $\frac{2}{x}\to0$. $$A=0$$