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.
 A: $(\frac{1}{2})^{2}$ does not converge to $0$ near $x = 0$, it converges to $1$. Hence you get $0$ in the denominator, also in the numerator. Thus the value of the limit cannot be deduced yet.
You can apply L'Hopital's rule to calculate the limit since in this case it is the $\frac{0}{0}$ symbol. Calculate the derivative of the numerator and then, separately, the derivative of the numerator. The limit at $x=0$ of the quotient of those two results is what you are looking for. Look up the geometrical interpretation of the rule.
A: If you know about L'hopital rule it may be easier:
$$\lim_{x\to0} \frac{e^{2x} - e^x}{4^x - 2^x}=\lim_{x\to0} \frac{2e^{2x} - e^x}{\log4\cdot4^x - \log 2\cdot2^x}=\frac{2-1}{2\log 2-\log2}=\frac{1}{\log 2}$$
A: If you know $\exp(x)=1+x+o(x)$ (which is consequence of $\exp'(x)|_{x=0}=1$), then using $a^x=e^{x\ln a}$:
$$\lim_{x\to 0}\frac{e^{2x}-e^{x}}{4^x-2^x}=\lim_{x\to 0}\frac{1+2x+o(2x)-1-x-o(x)}{1+x\ln 4 +o(x\ln 4)-1-x\ln 2-o(x\ln 2)}=\\
\lim_{x\to 0}\frac{2x-x+o(x)}{x(\ln 4-\ln 2)+o(x)}=\lim_{x\to 0}\frac{1+o(1)}{\ln 2+o(1)}=\frac{1}{\ln 2}$$
A: We can write:
$$\lim_{x\to 0}\frac{e^{2x}-e^x}{4^x-2^x}=\lim_{x\to 0}\frac{e^x\cdot(e^x-1)}{e^{\ln(2)x}\cdot(e^{\ln(2)x}-1)}=\lim_{x\to 0}\frac{e^x-1}{e^{(\ln(2)-1)x}\cdot (e^{\ln(2)x}-1)}\,\,\sim\,\,\lim_{x\to 0}\frac {x}{\ln(2)x}=\frac{1}{\ln(2)}$$
By remembering that:
$$e^{\alpha x}-1 \,\,\sim \,\, \alpha x$$
when $x\to 0$.
