Section 4.10 : L'Hospital's Rule and Indeterminate Forms
1. Use L’Hospital’s Rule to evaluate \(\displaystyle \mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 7{x^2} + 10x}}{{{x^2} + x - 6}}\).
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Start SolutionThe first step we should really do here is verify that L’Hospital’s Rule can in fact be used on this limit.
This may seem like a silly step given that we are told to use L’Hospital’s Rule. However, in later sections we won’t be told to use it when/if it can be used. Therefore, we really need to get in the habit of checking that it can be used before applying it just to make sure that we can. If we apply L’Hospital’s Rule to a problem that it can’t be applied to then it’s is almost assured that we will get the wrong answer (it’s always possible you might get lucky and get the correct answer, but we will only be very lucky if it does).
So, a quick check shows us that,
\[{\mbox{as}}\,\,\,x \to 2\hspace{0.5in}\frac{{{x^3} - 7{x^2} + 10x}}{{{x^2} + x - 6}} \to \frac{0}{0}\]and so this is a form that allows the use of L’Hospital’s Rule.
Show Step 2So, at this point let’s just apply L’Hospital’s Rule.
\[\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 7{x^2} + 10x}}{{{x^2} + x - 6}} = \mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - 14x + 10}}{{2x + 1}}\] Show Step 3At this point all we need to do is try the limit and see if it can be done.
\[\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 7{x^2} + 10x}}{{{x^2} + x - 6}} = \mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - 14x + 10}}{{2x + 1}} = \frac{{ - 6}}{5}\]So, the limit can be done, and we done with the problem! The limit is then,
\[\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 7{x^2} + 10x}}{{{x^2} + x - 6}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{6}{5}}}\]