<s>[INST] Simpler approach when calculating $\lim_{x\to0} \frac{e^{2x} - e^x}{4^x - 2^x}$? I know this is a really simple problem, but I can't figure out what logical mistake I am making on my approach :
$$\lim_{x\to0} \frac{e^{2x} - e^x}{4^x - 2^x}$$
Dividing numerator and denominator with $4^x$,
$$ \lim_{x\to0} \frac{({\frac{e^2}{4}) }^x-{(\frac{e}{4}})^x}{1 - ({\frac{1}{2}})^x} $$
$\left(\frac{1}{2}\right)^x$ and $\left(\frac{e}{4}\right)^x$ converge to $0,$ so I thought this limit converges to $1$.

(+ additional approach)
$$
    \lim_{x\to0} \frac{e^{2x} - e^x}{4^x - 2^x} \\
    = \lim_{x\to0} \frac{e^{2x} - e^x}{x} \cdot \frac{x}{4^x - 2^x} \\
    = \lim_{x\to0} \left(\frac{e^{2x} - 1}{x} - \frac{e^x - 1}{x} \right) \cdot \frac{1}{\left(\frac{4^x - 1}{x} - \frac{2^x - 1}{x} \right)}
$$
And this becomes $(2-1) \times \frac{1}{\ln4 - \ln2 } = \frac{1}{\ln2}$, using the definition of derivative.
I wonder if there's a simpler solution without using L'Hopital's rule.
 [/INST] If you know $\exp(x)=1+x+o(x)$ (which is consequence of $\exp'(x)|_{x=0}=1$), then using $a^x=e^{x\ln a}$:
$$\lim_{x\to 0}\frac{e^{2x}-e^{x}}{4^x-2^x}=\lim_{x\to 0}\frac{1+2x+o(2x)-1-x-o(x)}{1+x\ln 4 +o(x\ln 4)-1-x\ln 2-o(x\ln 2)}=\\
\lim_{x\to 0}\frac{2x-x+o(x)}{x(\ln 4-\ln 2)+o(x)}=\lim_{x\to 0}\frac{1+o(1)}{\ln 2+o(1)}=\frac{1}{\ln 2}$$
</s>