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

Question

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 1

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 $L=0$.

Answer 2

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, $\frac{x^{x^2}}{2^{2^x}}\lt\frac{1}{x}$.