limits - Prove $lim_{xtoinfty}frac{x^{x^2}}{2^{2^x}}$

How do you prove this formula

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

Since both top and bottom approaches infinity, I assume it is L'Hospital's rule to solve it, but after the first step I'm stuck
$$\lim_{x \to\infty}\frac{x^2 logx}{2^xlog2}$$
So how can I solve this problem, it seems the answer is infinity but I don't know how to approach that.

Answer

You can rewrite the limit as
$$\begin{aligned} L&=\lim_{x \to\infty}\frac{x^{x^2}}{2^{2^x}}\\ &=\lim_{x \to\infty}\frac{\exp\left(x^2\cdot\log x\right)}{\exp\left(2^x\cdot\log 2\right)}\\ &=\lim_{x \to\infty}\exp\left(x^2\cdot\log x-2^x\cdot\log 2\right) \end{aligned}$$

Since $\exp(.)$ is a continuous function, we can change the order to
$$L=\exp\left(\lim_{x\to\infty}\left(x^2\cdot\log x-2^x\cdot\log 2\right)\right)$$

and name the inner limit to $L'$
$$L'=\lim_{x\to\infty}\left(x^2\cdot\log x-2^x\cdot\log 2\right)$$

Note that $\log x

Obviously, the limit of the RHS expression as $x$ goes to $\infty$ is $-\infty$; because the growth of $x^3$ is not comparable with the exponential growth of $2^x$. Hence, $L'=-\infty$, which means $L=0$.