# Find the limit of $e^x/2^x$ as $x$ approaches infinity

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!

• Try using $a^x=e^{x\ln(a)}$ – nathan.j.mcdougall Jan 18 at 3:22
• $\frac{e}{2}\gt 1$ – John Douma Jan 18 at 3:28
• Your L'Hospital's rule approach actually works. Say the limit is $L$. You showed $L=\frac1{\ln2}L$, so the limit, if it exists, is zero. But it obviously isn't zero, so... – MJD Jan 18 at 3:32

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 $$\frac{e}{2} > 1$$.

• Yeah, it can't get any simpler than this approach. – Randall Jan 18 at 3:29

HINT

$$\frac{e^x}{2^x}=\frac{e^x}{e^{x\ln2}}=e^{x(1-\ln 2)}$$

Then:

$$\lim_{x\to\infty}{x(1-\ln 2)}= ?$$