hpfan101
  • hpfan101
How to find the horizontal and vertical asymptote of the function: \[h(x)=\frac{ \sqrt{2x^2+2} }{ x^2+-4 }\] using limits?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • 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 \(x\to\pm\infty\).
hpfan101
  • hpfan101
Alright, thank you for the explanation!

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