Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to\infty}\sqrt xe^{-x/2}=0$$
$$A=\lim_{x\to\infty}\sqrt xe^{-x/2}$$ $$A=\lim_{x\to\infty}\frac{\sqrt x}{e^{x/2}}$$ $$A=\lim_{x\to\infty}\frac{\sqrt x}{\sqrt{e^x}}$$ $$A=\lim_{x\to\infty}\sqrt{\frac{x}{e^x}}$$ $$A=\sqrt{\lim_{x\to\infty}\frac{x}{e^x}}$$ As $x\to\infty$, $e^x\to\infty$. We have an indeterminate form of $\infty/\infty$, which can dealt with by L'Hospital's Rule: $$A=\sqrt{\lim_{x\to\infty}\frac{x'}{(e^x)'}}$$ $$A=\sqrt{\lim_{x\to\infty}\frac{1}{e^x}}$$ We know from above that $\lim_{x\to\infty}(e^x)=\infty$, therefore, $\lim_{x\to\infty}\frac{1}{e^x}=0$. So, $$A=\sqrt0=0$$