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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?
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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Vertical 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 nonremovable 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 \(x\to\pm\infty\).

hpfan101
 one year ago
Best ResponseYou've already chosen the best response.0Alright, thank you for the explanation!
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