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

Question

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!

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 '19 at 03:32

score 5 · Accepted Answer · answered Jan 18 '19 at 03:28

5

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

answered Jan 18 '19 at 03:28

John Omielan

47,976

1

Yeah, it can't get any simpler than this approach. – Randall Jan 18 '19 at 03:29

score 4 · Answer 2 · answered Jan 18 '19 at 03:26

4

HINT

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

Then:

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

answered Jan 18 '19 at 03:26

Rhys Hughes

12,842

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

2 Answers2