Science:Math Exam Resources/Courses/MATH104/December 2013/Question 02 (f)

MATH104 December 2013

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Question 02 (f)
Consider the function $f(x)={\frac {x^{2}}{x^{2}-4}}.$ Its first and second derivatives are given by $f'(x)=-{\frac {8x}{(x^{2}-4)^{2}}},\quad f''(x)={\frac {8(3x^{2}+4)}{(x^{2}-4)^{3}}}$ (f) Find any horizontal and vertical asymptotes of the function $f(x)$ and write their equations.

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Hint
The function $f(x)$ has a vertical asymptote at $x=a$ if $\lim _{x\rightarrow a^{+}}f(x)=\infty \,\,{\text{or}}\,-\infty ,\quad {\text{and/or}}\quad \lim _{x\rightarrow a^{-}}f(x)=\infty \,\,{\text{or}}\,-\infty .$ The function $f(x)$ has a horizontal asymptote at $y=b$ if $\lim _{x\rightarrow +\infty }f(x)=b\quad {\text{and/or}}\quad \lim _{x\rightarrow -\infty }f(x)=b.$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Points where the numerator goes to infinity or the denominator goes to zero are candidates for where a fraction might have a vertical asymptote. In our case the denominator goes to zero for $\displaystyle x=\pm 2$. To evaluate the corresponding limits, plug in a value that is close to the limiting value on either side: $\displaystyle \lim _{x\to -2^{-}}{\frac {x^{2}}{x^{2}-4}}=\infty .$ The sign is positive since the numerator approaches 4 and plugging in a number slightly smaller than -2 yields a very small, positive denominator. Similar considerations yield ${\begin{aligned}\lim _{x\to -2^{+}}{\frac {x^{2}}{x^{2}-4}}&=-\infty \\\lim _{x\to 2^{-}}{\frac {x^{2}}{x^{2}-4}}&=-\infty \\\lim _{x\to 2^{+}}{\frac {x^{2}}{x^{2}-4}}&=\infty \end{aligned}}$ Thus there are vertical asymptotes at $\displaystyle {\color {blue}x=\pm 2}$. For the horizontal asymptotes, we check the limit of the function as $x$ goes to positive and negative infinity. We can do this by factoring out the largest power of the numerator and the denominator. $\displaystyle \lim _{x\to -\infty }{\frac {x^{2}}{x^{2}-4}}=\lim _{x\to -\infty }{\frac {x^{2}}{x^{2}(1-4/x^{2})}}=\lim _{x\to -\infty }{\frac {1}{1-4/x^{2}}}=1$ $\displaystyle \lim _{x\to \infty }{\frac {x^{2}}{x^{2}-4}}=\lim _{x\to \infty }{\frac {x^{2}}{x^{2}(1-4/x^{2})}}=\lim _{x\to \infty }{\frac {1}{1-4/x^{2}}}=1$ This tells us that we have horizontal asymptotes in both directions, at ${\color {blue}y=1}$.

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