Paul's Online Notes
Paul's Online Notes
Home / Calculus I / Applications of Derivatives / L'Hospital's Rule and Indeterminate Forms
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 4-10 : L'Hospital's Rule and Indeterminate Forms

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

Show All Steps Hide All Steps

Start Solution

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.

Show Step 2

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.

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