Asymptoottilaskin

Your input: find the vertical, horizontal and slant asymptotes of the function $$$f(x)=\frac{x^{2} + 2}{x^{2} - 4}$$$

The line $$$x=L$$$ is a vertical asymptote of the function $$$y=\frac{x^{2} + 2}{x^{2} - 4}$$$, if the limit of the function (one-sided) at this point is infinite.

In other words, it means that possible points are points where the denominator equals $$$0$$$ or doesn't exist.

So, find the points where the denominator equals $$$0$$$ and check them.

$$$x=-2$$$, check:

$$$\lim_{x \to -2^+}\left(\frac{x^{2} + 2}{x^{2} - 4}\right)=-\infty$$$ (for steps, see limit calculator).

Since the limit is infinite, then $$$x=-2$$$ is a vertical asymptote.

$$$x=2$$$, check:

$$$\lim_{x \to 2^+}\left(\frac{x^{2} + 2}{x^{2} - 4}\right)=\infty$$$ (for steps, see limit calculator).

Since the limit is infinite, then $$$x=2$$$ is a vertical asymptote.

Line $$$y=L$$$ is a horizontal asymptote of the function $$$y=f{\left(x \right)}$$$, if either $$$\lim_{x \to \infty} f{\left(x \right)}=L$$$ or $$$\lim_{x \to -\infty} f{\left(x \right)}=L$$$, and $$$L$$$ is finite.

Calculate the limits:

$$$\lim_{x \to \infty}\left(\frac{x^{2} + 2}{x^{2} - 4}\right)=1$$$ (for steps, see limit calculator).

$$$\lim_{x \to -\infty}\left(\frac{x^{2} + 2}{x^{2} - 4}\right)=1$$$ (for steps, see limit calculator).

Thus, the horizontal asymptote is $$$y=1$$$.

Since the degree of the numerator is not one degree greater than the denominator, then there are no slant asymptotes.

Answer

Vertical asymptotes: $$$x=-2$$$; $$$x=2$$$

Horizontal asymptote: $$$y=1$$$

No slant asymptotes.