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.
Can someone please help?
$\endgroup$ 21 Answer
$\begingroup$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.
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