Prove $\lim_{x\to\infty}\frac{x^{x^2}}{2^{2^x}}$ 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.
 A: We have $x^k\lt2^x$ for any $k$ and all large enough $x$. To show this we use the definition:
$$
e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}
$$
Hence, $\frac{x^{k+1}}{(k+1)!}\lt e^x $, so that $\frac{x^k}{e^x} \lt\frac{(k+1)!}{x}\rightarrow0$ as ${x}\rightarrow\infty$.
Similarily we see that $\log x \lt x$ for all large enough $x$. Thus for all large enough $x$:
$$
\begin{equation}
\begin{split}
\log x\cdot\left(x^2+1\right) &\lt x^3\lt\log 2 \cdot2^x \\
\exp\left(\log x\cdot\left(x^2+1\right) \right) &\lt\exp\left(\log 2\cdot 2^x\right) \\
x^{x^2+1} &\lt 2^{2^x} \\
\end{split}
\end{equation}
$$
Hence, $\frac{x^{x^2}}{2^{2^x}}\lt\frac{1}{x}$.
A: 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 $\log x<x$, so $$x^2\log x<x^3$$ and therefore $$x^2\cdot\log x-2^x\cdot\log 2 < x^3-2^x\cdot\log 2$$ By taking $x$ to infinity we'll get $$L'<\lim_{x\to\infty}\left(x^3-2^x\cdot\log 2\right)$$
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 $L=0$.
