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

Question

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.

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

Answer 1

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?

Answer 2

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

Answer 3

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*}