How to evaluate this limit without using L'Hospital's rule?

How do I solve this limit without using the l'Hospital's rule? For whatever strange reason, my teacher wants this done without the l'Hospital's rule.

$$\lim_{x\to 5^-}\frac{e^x}{(x-5)^3}.$$

• Could you please reformat your limit in Latex? It's very ambiguous as typed now. I wanted to edit it, but couldn't be sure I would get the correct expression. – Deepak Oct 3 '14 at 3:55
• Actually, L'Hospital's rule cannot be used for this limit... L'Hospital's rule is not a magic trick that always save the day, there are conditions to apply it! – Taladris Oct 3 '14 at 4:23

Here's something to keep in mind: what happens to $e^x$ as $x\to 5$? Is this number zero or not? What happens to the denominator?

• If this (your edit) is the intended limit, then L' Hopital's Rule wouldn't even be applicable in the first place. So why should the teacher explicitly proscribe its use? I would prefer to let the asker confirm if this is the intended limit. – Deepak Oct 3 '14 at 4:05
• Yes this is the intended limit. – Jason Browns Oct 3 '14 at 4:20
• If it's the intended limit, LHR would not be applicable and it's actually a fairly obvious limit. The numerator is finite and positive, the denominator is zero but approaching it from the left side, so...? – Deepak Oct 3 '14 at 4:23
• Well just by looking at it I am pretty sure the answer would be -infinity, but the problem is I don't know to show work for that. – Jason Browns Oct 3 '14 at 4:28

Since $x\lt 5$ as $x\to 5^-$, then the denominator is negative. Therefore $$\lim_{x\to 5^-} \frac{e^x}{(x-5)^3} =-\infty$$

Observe that we can split the limit up giving us the following:

\begin{equation*} \lim_{x\to 5^-}e^x\lim_{x\to 5^-}\frac{1}{(x-5)^3}. \end{equation*}

The first limit approaches $e^5$ as $x\to 5^{-}$ and for the second limit, $(x-5)^3$ approaches zero as $x\to 5^{-}$so $\frac{1}{(x-5)^3}$ goes to $-\infty.$ Therefore,

\begin{equation*} \lim_{x\to 5^-} \frac{e^x}{(x-5)^3}=-\infty.~_{\square} \end{equation*}