Finding the horizontal asymptotes of a function

Question

Here is the function I am trying to find the horizontal asymptotes for:

$y=\frac{1-2^x}{1+2^x}$

Could you please explain how the horizontal asymptotes can be found for this function?

Thanks.

Answer 1

"Horizontal asymptotes" is a geometric statement of the limits of the following two limits: $$\lim_{x \to +\infty} \frac{1 - 2^x}{1 + 2^x} \text{ and } \lim_{x \to -\infty}\frac{1 - 2^x}{1 + 2^x}.$$ in which \begin{align*} & \lim_{x \to +\infty} \frac{1 - 2^x}{1 + 2^x} = \lim_{x \to +\infty}\frac{\frac{1}{2^x} - 1}{\frac{1}{2^x} + 1} = \frac{0 - 1}{0 + 1} = -1, \\ & \lim_{x \to -\infty}\frac{1 - 2^x}{1 + 2^x} = \frac{1 - 0}{1 + 0} = 1. \end{align*} Hence the two horizontal asymptotes are straight lines $y = -1$ and $y = 1$.