Section 5.5 # 5

Section 5.5 # 5

Matthew Sweeney 07 Nov 2010 20:50

When sketching this function, I determined that there's a horizontal asymptote at 0 but only by intuition and not by any mathematical means. I know that to find horizontal asymptotes, you must find $\displaystyle\lim_{x \to \infty}f(x)$ and $\displaystyle\lim_{x \to -\infty}f(x)$. What I'm guessing is that with $\displaystyle\lim_{x \to \infty}f(x)$, you basically get $(\infty)(\infty)$, meaning it shoots off up to infinity, but with $\displaystyle\lim_{x \to -\infty}f(x)$, it would be something like $\frac{\infty}{\infty}$, with the bottom growing faster than the top. Does this intuitively mean that as $x \to \infty$, $f(x) \to 0$? If not, how is this determined? I feel like I should know this but for some reason the answer isn't coming to mind.

Re: Section 5.5 # 5

Dana Ernst 07 Nov 2010 21:09

You have the right idea, but with the limit as $x\to -\infty$, you intitially have the indeterminate form $-\infty\cdot 0$. The basic idea is to rewrite and use L'Hospital's rule:

(1)

\begin{align} \lim_{x\to -\infty}xe^x & = \lim_{x\to -\infty}\frac{x}{e^{-x}}\\ & \stackrel{\text{L'H}}{=} \lim_{x\to -\infty} \frac{1}{-e^{-x}}\\ & = \lim_{x\to -\infty} -e^x\\ & = 0 \end{align}

Therefore, there is a horizontal asymptote at $y=0$ and the graph approaches it on the left, not right.