calculus - How to find $lim_{xtoinfty}{frac{e^x}{x^a}}$?

$$\lim_{x\to\infty}{\frac{e^x}{x^a}}$$ For $a\in \Bbb R,$ find this limit.

I would say for $a\ge 0$ lim is equal to $\lim_{x\to\infty}{\frac{e^x}{a!x^0}=\infty}$ (from L'Hopital).

For $a<0$, lim eq. to $\frac{\infty}{0}$so lim doesnt exist. Is this correct?

Answer

Because $e^x > x$ for all $x$, $$\lim_{x \to \infty}\frac{e^x}{x}=\lim_{x \to \infty}\frac{1}{2}\left(\frac{e^{x/2}}{x/2}\right)e^{x/2} = \infty.$$ since $e^{x/2}/(x/2) > 1,$ and $e^{x/2} \to \infty$ as $x \to \infty$.

Then, it follows that $$\lim_{x \to \infty}\frac{e^x}{x^a}=\lim_{x \to \infty}\frac{1}{a^a}\cdot \left( \frac{e^{x/a}}{x/a}\right)^a = \infty$$

since we just showed that what is in parentheses approaches $\infty$ as $x \to \infty$, so the whole limit has to go to $\infty.$