Solving $\lim_{x\to\infty} \frac{x^{2x}}{x^{5x}}$

Question

I'm trying to solve the expression $$\lim_{x\to\infty} \frac{x^{2x}}{x^{5x}}.$$ I know the limit is equal to $0,$ and since the numerator and denominator go to infinity as $x$ goes to infinity, I should be able to use L'Hôpital's rule but it doesn't seem to work.

Is this the wrong approach? If so, how would you solve this?

Answer 1

$$\frac{x^{2x}}{x^{5x}}=x^{-3x}=\frac{1}{(x^x)^3}$$ No need for L'Hopital's rule, as the denominator goes to $\infty $ as $x\to\infty$.

The question in the title is different, but solved the same way.

Answer 2

$$\frac {x^{2x}}{x^{5x}}=x^{2x-5x}=x^{-3x} $$

$$=e^{-3x\ln (x)} $$

When $x\to +\infty, -3x\ln (x)\to-\infty $ and

$$e^{-3x\ln (x)}\to 0$$