L'Hospital's Rule Examples for Indeterminate Quotients

Recall from the Indeterminate Forms page:

Sometimes we have limits in the form of $\lim_{x\rightarrow a} \frac{f(x)}{g(x)}$ where as $x \to a$, $f(x) \to 0$ and $g(x) \to 0$. Thus we have a limit of Indeterminate Form of Type $\frac{0}{0}$.

For limits in the form $\lim_{x\rightarrow a} \frac{f(x)}{g(x)}$ where as $x \to a$, $f(x) \to \pm \infty$ and $g(x) \to \pm \infty$. Thus we have a limit of Indeterminate Form of Type $\frac{\infty}{\infty}$.

Now let's try out some examples using L'Hospital's rule.

Example 1

Evaluate the following limit: $\lim_{t \to 0} \frac{e^{2t} - 1}{\sin t}$.

Notice that as $t \to 0$, $e^{2t} - 1 \to 0$ and $\sin t \to 0$. Thus we can apply L'Hospital's rule:

(1)

\begin{align} \lim_{t \to 0} \frac{e^{2t} - 1}{\sin t} \overset{H} = \lim_{t \to 0} \frac{2e^{2t}}{\cos t} = 2 \end{align}

Example 2

Evaluate the following limit: $\lim_{t \to 1} \frac{t^6 - 1}{t^5 - 1}$.

As $t \to 1$, $t^6 - 1 \to 1$ and $t^5 - 1 \to 1$, so we can use L'Hospital's rule:

(2)

\begin{align} \lim_{t \to 1} \frac{t^6 - 1}{t^5 - 1} \overset{H} = \lim_{t \to 1} \frac{6t^5}{5t^4} = \lim_{t \to 1} \frac{6t}{5} = \frac{6}{5} \end{align}

Example 3

Evaluate the following limit: $\lim_{x \to 0} \frac{x 3^x}{3^x - 1}$.

As $x \to 0$, $x 3^x \to 0$, and $3^x - 1 \to 0$. Let's use L'Hospital's rule:

(3)

\begin{align} \lim_{x \to 0} \frac{x 3^x}{3^x - 1} \overset{H} = \lim_{x \to 0} \frac{3^x + x \cdot 3^x \ln 3}{3^x \ln 3} = \frac{1}{\ln 3} \end{align}