I am trying to find the asymptotic relation between $e^x$ and $2^x$. I tried to use limit comparison: $$\lim_{x\to\infty}\left(\frac{e^x}{2^x}\right)$$
I tried to use L'Hopital's rule:
$$\lim_{x\to\infty}\left(\frac{e^x}{2^x}\right)$$
$$=\lim_{x\to\infty}\left(\frac{e^x}{\ln(2) \cdot 2^x}\right)$$
which doesn't really help. Is there any way to compute this limit? Thanks!
Best Answer
Note that
$$\lim_{x \to \infty} \left(\cfrac{e^x}{2^x}\right) = \lim_{x \to \infty} \left(\cfrac{e}{2}\right)^x = \infty \tag{1}\label{eq1}$$
because $e > 2$ so $\frac{e}{2} > 1$.