## hpfan101 one year ago How to find the horizontal and vertical asymptote of the function: $h(x)=\frac{ \sqrt{2x^2+2} }{ x^2+-4 }$ using limits?

1. anonymous

Vertical asymptotes occur for values of $$x$$ that make the given function approach positive/negative infinity from either direction (i.e. the function approaches $$+\infty$$, $$-\infty$$, or the limit doesn't exist because the function approaches $$\infty$$ from one side and $$-\infty$$ from the other). In other words, if $\lim_{x\to c}f(x)=\pm\infty\quad\text{or}\quad \begin{matrix}\lim\limits_{x\to c^-}f(x)=\pm\infty\\\text{and}\\ \lim\limits_{x\to c^+}f(x)=\mp \infty\end{matrix}$ then a vertical asymptote is said to occur at $$x=c$$. Protip: for rational functions such as yours, this happens when the denominator approaches $$0$$ and the numerator does not (i.e. if $$h(x)$$ has a non-removable singularity at $$x=c$$). Oblique asymptotes are similar but different - Instead of the function approaching an infinite value, the independent variable does. A polynomial $$p(x)$$ is an oblique asymptote of $$f(x)$$ if $\lim_{x\to\pm\infty}\bigg[f(x)-p(x)\bigg]=0$ In the simplest case, $$p(x)$$ is a constant, and so the asymptote would be horizontal. All this to say you can find horizontal asymptotes by checking the limit as $$x\to\pm\infty$$.

2. hpfan101

Alright, thank you for the explanation!