# Finding the vertical and horizontal asymptotes of the function $f(x) = \frac{e^x(x + 1)}{e^{2x}(x^2 - 1)}$

I am to find the vertical and horizontal asymptotes of this given function:

$$f(x) = \frac{e^x(x + 1)}{e^{2x}(x^2 - 1)}$$

To find the vertical asymptote,I think I equate the bottom line to zero and whatever my $$x$$ gives is the vertical asymptote? But I don't know how to solve $$e^{2x}(x^2-1) = 0$$. Is it going to be $$x = +1,-1$$? Hence, the vertical asymptotes are $$1$$ and $$-1$$.

To find horizontal asymptotes, I think I should find the limit as $$x \to \infty$$ and $$x \to -\infty$$. I don't know how to go about this.

• You should first simplify to get $f(x) = \frac{1}{e^x(x-1)}$ then you see that the vertical asymptote is only at $x=1$. The horizontal asymptote is answered below. Commented Apr 4, 2022 at 17:05
• This tutorial explains how to typeset mathematics on this site. Commented Apr 6, 2022 at 11:36

You can just calculate the value of $$y$$ at $$x\to-\infty \;\text{and} \; x \to\infty$$ $$\lim_{x\to-\infty}\frac{e^x(x+1)}{e^{2x}(x^2-1)}=-\infty$$ $$\lim_{x\to\infty}\frac{e^x(x+1)}{e^{2x}(x^2-1)}=0$$

As limit exists for $$x\to\infty$$, $$y=0$$ is the horizontal asymptote.

For vertical asymptote, you must find values of $$x$$ for which $$y\to\infty$$.

$$y=f(x)=\frac{e^x(x+1)}{e^{2x}(x^2-1)}$$

As denominator goes to $$0$$ for $$x=\pm1$$, we will check limit of $$y$$ at these values.

$$\lim_{x\to-1}\frac{e^x(x+1)}{e^{2x}(x^2-1)}=-\frac{e}{2}$$ $$\lim_{x\to1}\frac{e^x(x+1)}{e^{2x}(x^2-1)}=\infty$$

Since $$y \to\infty$$ for $$x=1$$, $$x=1$$ is the vertical asymptote.

$$x=-1$$, is not the vertical asymptote as limit of $$y$$ is defined at this value.

• So the vertical asymptotes are +1 and -1 right? Commented Apr 4, 2022 at 17:07
• @yowhatsup123, no only $x=1$ is the vertical asymptote. Commented Apr 4, 2022 at 17:08