Limit of $\frac{e^{x}}{\ln(x)}$

Question

I don't know how to find the limit $$\lim_{x\to\infty}\frac{e^x}{\log x}.$$ How can I do this ?

Thank you in advance.

Answer 1

For $x > 1$, $$ \frac{e^x}{\ln x} > \frac{1 + x + \frac{x^2}{2!}}{x} > \frac{x}{2} \to \infty. $$ The relevant inequalities can be proved by elementary means; see for example here and here.

Answer 2

You may also use the L'Hôpital's rule which, in this case, in view of the answer given by Goos, would be like killing a fly with a cannon...

I am giving you this information because L'Hôpital's rule is quite useful in some much more difficult problems involving limits (see the link for such examples), so it's good to know it.

Answer 3

$$e^x>x\log x,\quad x\ge 1,$$ since $e=e^1<1\cdot\log 1=0$ and $$e^x=\frac{d}{dx}e^x<\frac{d}{dx}(x\log x)=1+\log x.$$

Answer 4

$\frac{\exp(x)}{\ln(x)} > \frac{\exp(x)}{x} > x$

First part because $x > \ln(x)$

Second part because $\exp(x) > x^2$ as $x^2 = \exp(2 \cdot \ln(x))$ and $x > 2 \cdot \ln(x)$ for $x$ greater than 2. And exponential is an increasing function