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