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.

If you know about L'hopital rule it may be easier:
$$\lim_{x\to0} \frac{e^{2x} - e^x}{4^x - 2^x}=\lim_{x\to0} \frac{2e^{2x} - e^x}{\log4\cdot4^x - \log 2\cdot2^x}=\frac{2-1}{2\log 2-\log2}=\frac{1}{\log 2}$$
