Answer
$\lim\limits_{x \to 0}\frac{e^{2x}-e^{-2x}}{ln(x+1)} = 4$
Work Step by Step
$\lim\limits_{x \to 0}\frac{e^{2x}-e^{-2x}}{ln(x+1)} = \frac{0}{0}$
We can use L'Hospital's Rule:
$\lim\limits_{x \to 0}\frac{e^{2x}-e^{-2x}}{ln(x+1)}$
$=\lim\limits_{x \to 0}\frac{2e^{2x}+2e^{-2x}}{1/(x+1)}$
$=\frac{2(1)+2(1)}{1}$
$= 4$
