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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10. Use L’Hospital’s Rule to evaluate \(\mathop {\lim }\limits_{y \to {0^ + }} {\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}}\).

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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 certain classes of rational functions and this is clearly not a rational function.

Note however that it is in the following indeterminate form,

\[{\mbox{as}}\,\,\,y \to {0^ + }\hspace{0.5in}{\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}} \to {1^\infty }\]

and as we discussed in the notes for this section we can do some manipulation on this to turn it into a problem that can be done with L’Hospital’s Rule.

Show Step 2

First, let’s define,

\[z = {\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}}\]

and take the log of both sides. We’ll also do a little simplification.

\[\ln z = \ln \left( {{{\left[ {\cos \left( {2y} \right)} \right]}^{{}^{1}/{}_{{{y^{\,2}}}}}}} \right) = \frac{1}{{{y^{\,2}}}}\ln \left[ {\cos \left( {2y} \right)} \right] = \frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}}\] Show Step 3

We can now take the limit as \(y \to {0^ + }\) of this.

\[\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \left[ {\frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}}} \right]\]

Before we proceed let’s notice that we have the following,

\[{\mbox{as}}\,\,\,y \to {0^ + }\hspace{0.5in}\frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}} \to \frac{{\ln \left( 1 \right)}}{0} = \frac{0}{0}\]

and we have a limit that we can use L’Hospital’s Rule on.

Show Step 4

Applying L’Hospital’s Rule gives,

\[\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \left[ {\frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}}} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{{}^{{ - 2\sin \left( {2y} \right)}}/{}_{{\cos \left( {2y} \right)}}}}{{2y}} = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{ - \tan \left( {2y} \right)}}{y}\] Show Step 5

We now have a limit that behaves like,

\[{\mbox{as}}\,\,\,y \to {0^ + }\hspace{0.5in}\frac{{ - \tan \left( {2y} \right)}}{y} \to \frac{0}{0}\]

and so we can use L’Hospital’s Rule on this as well. Doing this gives,

\[\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{ - \tan \left( {2y} \right)}}{y} = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{ - 2{{\sec }^2}\left( {2y} \right)}}{1} = - 2\] Show Step 6

Now all we need to do is recall that,

\[z = {{\bf{e}}^{\ln z}}\]

This in turn means that we can do the original limit as follows,

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