Science:Math Exam Resources/Courses/MATH104/December 2014/Question 05 (e)/Solution 1

We have already determined what the vertical asymptote is in part $a$, by seeing where $f$ is undefined. To see the behaviour of $f$ as it approaches $x=0$ from the right and left, we have the following calculations:

${\begin{aligned}\displaystyle \lim _{x\rightarrow 0^{+}}{\frac {e^{x}}{x^{2}}}&={\frac {\displaystyle \lim _{x\rightarrow 0^{+}}e^{x}}{\displaystyle \lim _{x\rightarrow 0^{+}}x^{2}}}\\&={\frac {1}{\displaystyle \lim _{x\rightarrow 0^{+}}x^{2}}}=\infty \end{aligned}}$

${\begin{aligned}\lim _{x\rightarrow 0^{-}}{\frac {e^{x}}{x^{2}}}&={\frac {\displaystyle \lim _{x\rightarrow 0^{-}}e^{x}}{\displaystyle \lim _{x\rightarrow 0^{-}}x^{2}}}\\&={\frac {1}{\displaystyle \lim _{x\rightarrow 0^{-}}x^{2}}}=\infty \end{aligned}}$

where we obtain the last equality since the square of any negative number is a positive number. This happens when $x=0$ so that is our vertical asymptote. To find the horizontal asymptote, we find $\displaystyle \lim _{x\rightarrow \infty }f(x)$ and $\displaystyle \lim _{x\rightarrow -\infty }f(x)$. Essentially, what we are determining is the behaviour of $f(x)$ as $x$ gets very large and very small. We will apply L’Hopital’s rule when finding the two limits.

${\begin{aligned}\lim _{x\rightarrow \infty }{\frac {e^{x}}{x^{2}}}&=\lim _{x\rightarrow \infty }{\frac {e^{x}}{2x}}\\&={\frac {1}{2}}\lim _{x\rightarrow \infty }{\frac {e^{x}}{x}}\end{aligned}}$

Since $\displaystyle \lim _{x\rightarrow \infty }{\frac {e^{x}}{x}}=\infty$, we have that $\displaystyle \lim _{x\rightarrow \infty }{\frac {e^{x}}{x^{2}}}=\infty$ (the limit does not exist). So as $x$ gets very large, $f(x)$ also gets very large and proceeds towards $\infty$.

${\begin{aligned}\lim _{x\rightarrow -\infty }{\frac {e^{x}}{x^{2}}}&=\lim _{x\rightarrow -\infty }{\frac {e^{x}}{2x}}\\&={\frac {1}{2}}\lim _{x\rightarrow -\infty }{\frac {e^{x}}{x}}\\&={\frac {1}{2}}\lim _{x\rightarrow -\infty }{\frac {e^{x}}{1}}\\&=0\end{aligned}}$

In the second limit calculation, we applied L’Hopital’s rule twice. We learn from the second calculation that as $x$ gets very small and speeds towards $-\infty$, $f(x)$ approaches $0$.