I need help with finsing the asymptotes of the graph (1 + ex)/(1 − ex).
Would they be a vertical one at 0, and 2 horizontal at 1 and -1? All of this is just based on me looking at the plotted graph. Thanks
Vertical asymptote: Solve $\displaystyle 1 - e^x = 0: x = 0$. So the line x = 0 (that is, the y-axis) is a vertical asymptote.
Horizontal asymptotes:
$\displaystyle \lim_{x \rightarrow -\infty} \frac{1+e^x}{1 - e^x} = \frac{1}{1} = 1$.
$\displaystyle \lim_{x \rightarrow +\infty} \frac{1+e^x}{1 - e^x} = \lim_{x \rightarrow +\infty} \frac{e^{-x} + 1}{e^{-x} - 1} = \frac{1}{-1} = -1$.
So the lines y = 1 and y = -1 are horizontal asymptotes.