Answer
Answer: 1
Work Step by Step
$\lim\limits_{x \to \infty} \frac{e^x + e^{-x}}{e^x- e^{-x}}$
First, we simply this limit.
$\frac{e^x + e^{-x}}{e^x- e^{-x}} = \frac{e^x + \frac{1}{e^{x}}}{e^x - \frac{1}{e^{x}}}$ (Because $e^{-x} = \frac{1}{e^x}$)
Thus, we have: $\frac{\frac{e^{2x} + 1}{e^x}}{\frac{e^{2x} - 1}{e^x}} = \frac{e^{2x} + 1}{e^{2x} - 1}$.
Dividing both numerator and denominator by $e^{2x}$, we have: $\frac{1 +\frac{1}{e^{2x}}}{1 -\frac{1}{e^{2x}}}$.
We know that
$\lim\limits_{x \to \infty} e^{x} = \infty$
Hence, $\lim\limits_{x \to \infty} e^{2x} = \infty$
Hence, Hence, $\lim\limits_{x \to \infty} \frac{1}{e^{2x}} = 0$.
Thus, $\lim\limits_{x \to \infty} \frac{1 +\frac{1}{e^{2x}}}{1 -\frac{1}{e^{2x}}} = \frac{1 + 0}{1 - 0} = 1$
