L'Hospital's Rule Examples for Indeterminate Products

Recall from the Indeterminate Forms page that:

If we have a limit in the form $\lim_{x\rightarrow a} f(x)g(x)$, that is as $x \to a$, $f(x) \to 0$ and $g(x) \to \pm \infty$ (or the other way around), then we have a limit of Indeterminate Form of Type $0 \times \infty$.

Of course, we can rewrite this function that we are taking the limit of as $f(x)g(x) = \frac{f(x)}{\frac{1}{g(x)}} = \frac{g(x)}{\frac{1}{f(x)}}$.

We will now look at some examples.

Example 1

Evaluate the following limit $\lim_{x \to 0} \cot 2x \cdot \sin 6x$.

Notice that as $x \to 0$, $\cot 2x \to \pm \infty$, and $\sin 6x \to 0$. Hence we can apply L'Hospital's rule after we rewrite the limit as $\lim_{x \to 0} \frac{\sin 6x}{\tan 2x}$

(1)

\begin{align} \lim_{x \to 0} \frac{\sin 6x}{\tan 2x} \overset{H} = \lim_{x \to 0} \frac{6 \cos x}{2 \sec ^2 2x} = \lim_{x \to 0} \frac{6\cos 6x \cdot \cos ^2 2x}{2} = 3 \end{align}

Example 2

Evaluate the limit $\lim_{x \to 1^+} \ln x \cdot \tan \left ( \frac{\pi x}{2} \right)$.

We can apply L'Hospital's rule if we rewrite this limit to get an indeterminate quotient:

(2)

\begin{align} \lim_{x \to 1^+} \ln x \cdot \tan (\frac{\pi x}{2}) = \lim_{x \to 1^+} \frac{\ln x}{\frac{1}{\tan (\frac{\pi x}{2})}} \end{align}

Now let's use L'Hospital's rule:

(3)

\begin{align} \quad \quad \lim_{x \to 1^+} \frac{\ln x}{\frac{1}{\tan \left (\frac{\pi x}{2} \right)}} \overset{H} = \lim_{x \to 1^+} \frac{\frac{1}{x}}{ \left (\frac{{\frac{\pi}{2} \sec ^2 \left (\frac{\pi x}{2} \right )}}{ \left (\tan \left (\frac{\pi x}{2}\right ) \right)^2 }\right)} = \lim_{x \to 1^+} \frac{2 \left (\tan \left (\frac{\pi x}{2}\right ) \right)^2 \cdot \cos^2 \left (\frac{\pi x}{2} \right)}{\pi x} = 0 \end{align}