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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.

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

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  • $\begingroup$ 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. $\endgroup$
    – Deepak
    Oct 3, 2014 at 3:55
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    $\begingroup$ 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! $\endgroup$
    – Taladris
    Oct 3, 2014 at 4:23

3 Answers 3

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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?

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  • $\begingroup$ 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. $\endgroup$
    – Deepak
    Oct 3, 2014 at 4:05
  • $\begingroup$ Yes this is the intended limit. $\endgroup$ Oct 3, 2014 at 4:20
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    $\begingroup$ 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...? $\endgroup$
    – Deepak
    Oct 3, 2014 at 4:23
  • $\begingroup$ 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. $\endgroup$ Oct 3, 2014 at 4:28
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Since $x\lt 5$ as $x\to 5^-$, then the denominator is negative. Therefore $$ \lim_{x\to 5^-} \frac{e^x}{(x-5)^3} =-\infty $$

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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*}

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