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

8. Use L’Hospital’s Rule to evaluate \(\mathop {\lim }\limits_{w \to {0^ + }} \left[ {{w^2}\ln \left( {4{w^2}} \right)} \right]\).

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The first thing to notice here is that is not in a form that allows L’Hospital’s Rule. L’Hospital’s Rule only works on a certain class of rational functions and this is clearly not a rational function.

Note however that it is in the following indeterminate form,

\[{\mbox{as}}\,\,\,w \to {0^ + }\hspace{0.5in}{w^2}\ln \left( {4{w^2}} \right) \to \left( 0 \right)\left( { - \infty } \right)\]

and as we discussed in the notes for this section we can always turn this kind of indeterminate form into a rational expression that will allow L’Hospital’s Rule to be applied.

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The real question is do we move the first term or the second term to the denominator. From the looks of things, it appears that it would be best to move the first term to the denominator.

\[\mathop {\lim }\limits_{w \to {0^ + }} \left[ {{w^2}\ln \left( {4{w^2}} \right)} \right] = \mathop {\lim }\limits_{w \to {0^ + }} \frac{{\ln \left( {4{w^2}} \right)}}{{{}^{1}/{}_{{{w^2}}}}}\]

Notice as well that,

\[{\mbox{as}}\,\,\,w \to {0^ + }\hspace{0.5in}\frac{{\ln \left( {4{w^2}} \right)}}{{{}^{1}/{}_{{{w^2}}}}} \to \frac{{ - \infty }}{\infty }\]

and we can use L’Hospital’s Rule on this.

Show Step 3

Applying L’Hospital’s Rule gives,

\[\mathop {\lim }\limits_{w \to {0^ + }} \left[ {{w^2}\ln \left( {4{w^2}} \right)} \right] = \mathop {\lim }\limits_{w \to {0^ + }} \frac{{\ln \left( {4{w^2}} \right)}}{{{}^{1}/{}_{{{w^2}}}}} = \mathop {\lim }\limits_{w \to {0^ + }} \frac{{{}^{2}/{}_{w}}}{{ - {}^{2}/{}_{{{w^3}}}}}\]

Can you see why we chose to move the first term to the denominator? Moving the logarithm would have left us with a very messy derivative to take! It might have ended up working okay for us, but the work would be greatly increased.

Show Step 4

Do not forget to simplify after we’ve taken the derivative. This problem becomes very simple if we do that. In fact, it is the only way to actually get an answer for this problem. If we do not simplify will get stuck in a never-ending chain of infinity divided by infinity forms no matter how many times we apply L’Hospital’s Rule.

\[\mathop {\lim }\limits_{w \to {0^ + }} \left[ {{w^2}\ln \left( {4{w^2}} \right)} \right] = \mathop {\lim }\limits_{w \to {0^ + }} \frac{{\ln \left( {4{w^2}} \right)}}{{{}^{1}/{}_{{{w^2}}}}} = \mathop {\lim }\limits_{w \to {0^ + }} \left( { - {w^2}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{0}\]