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!

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    $\begingroup$ Try using $a^x=e^{x\ln(a)}$ $\endgroup$ – nathan.j.mcdougall Jan 18 '19 at 3:22
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    $\begingroup$ $\frac{e}{2}\gt 1$ $\endgroup$ – John Douma Jan 18 '19 at 3:28
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    $\begingroup$ 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... $\endgroup$ – MJD Jan 18 '19 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 $\frac{e}{2} > 1$.

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    $\begingroup$ Yeah, it can't get any simpler than this approach. $\endgroup$ – Randall Jan 18 '19 at 3:29


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


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


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