Answer
$1$
Work Step by Step
Find the limit$\lim\limits_{x \to \infty}\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$.
Thus, $\lim\limits_{x \to \infty}\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}=\lim\limits_{x \to \infty}\frac{e^{3x}(1-e^{-6x})}{e^{3x}(1+e^{-6x})}$
$=\frac{1-\lim\limits_{x \to \infty }e^{-6x}}{1+\lim\limits_{x \to \infty} e^{-6x}}$
$=\frac{1-0}{1+0}$
$=1$
