Calculus I - L'Hospital's Rule and Indeterminate Forms

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Section 4.10 : L'Hospital's Rule and Indeterminate Forms

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4. Use L’Hospital’s Rule to evaluate \(\displaystyle \mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}}\).

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The 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,

\[{\mbox{as}}\,\,\,z \to 0\hspace{0.5in}\frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}} \to \frac{0}{0}\]

and so this is a form that allows the use of L’Hospital’s Rule.

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Before actually using L’Hospital’s Rule it might be better if we multiply out the denominator to make the derivative (and later steps a little easier). Doing this gives,

\[\mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}} = \mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^4} + 2{z^3} + {z^2}}}\]

Now let’s apply L’Hospitals’s Rule.

\[\mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}} = \mathop {\lim }\limits_{z \to 0} \frac{{2\cos \left( {2z} \right) + 14z - 2}}{{4{z^3} + 6{z^2} + 2z}}\] Show Step 3

At this point let’s try the limit and see if it can be done. However, in this case, we can see that,

\[{\mbox{as}}\,\,\,z \to 0\hspace{0.5in}\frac{{2\cos \left( {2z} \right) + 14z - 2}}{{4{z^3} + 6{z^2} + 2z}} \to \frac{0}{0}\] Show Step 4

So, using L’Hospital’s Rule doesn’t give us a limit that we can do. However, the new limit is one that can use L’Hospital’s Rule on so let’s do that.

\[\mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}} = \mathop {\lim }\limits_{z \to 0} \frac{{ - 4\sin \left( {2z} \right) + 14}}{{12{z^2} + 12z + 2}} = \frac{{14}}{2}\]

Okay, the second L’Hospital’s Rule gives us a limit we can do and so the answer is,

\[\mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}} = \require{bbox} \bbox[2pt,border:1px solid black]{7}\]