How to compute $\lim_{x\to\infty} \frac{2^x}{x^{200}}$?

Question

I am trying to solve this limit: $$\lim_{x\to\infty} \frac{2^x}{x^{200}}$$

The first result is an indetermination of the kind $\frac{\infty}{\infty}$ but here applying L'Hopital would be too long, I do not see any substitution by means of equivalent infinitesimals possible and simplifying the limit neither.

How can I solve it? The solution must be $\infty$. Thank you!

Answer 1

L'Hopital rule applied $200$ times leads to $$\lim_{x\to\infty}\frac{(\log 2)^{200} \,2^x}{200!}=\infty$$

Or you can take the logarithm of the limit

$$\log\lim_{x\to\infty}\frac{2^x}{x^{200}}=\lim_{x\to\infty}\left(x\log 2-200\log x\right)=\lim_{x\to\infty}x\left(\log 2-200\frac{\log x}{x}\right)=\infty$$

As the log of the limit is $+\infty$ the limit is $+\infty$

Answer 2

$$\lim_{x \to \infty} \frac{2^x}{x^{200}} = \lim_{x \to \infty} \frac{2^x}{2^{200\log_2(x)}} = \lim_{x \to \infty} 2^{x-200 \log_2(x)} = 2^{\lim_{x \to \infty}x-200 \log_2(x)}$$

That limit clearly goes to infinity, so the whole function does as well

Answer 3

$$ \lim_{x\to\infty} \frac{2^x}{x^{200}}=\lim_{x\to\infty} \frac{e^{x\ln2}}{x^{200}}\rightarrow \mbox{(L'Hopital 200 times)}=\lim_{x\to\infty} \frac{(\ln2)^{(200)}e^{x\ln2}}{200!}=\infty $$