L'Hospital's Rule Examples for Indeterminate Differences

# L'Hospital's Rule Examples for Indeterminate Differences

Recall from the Indeterminate Forms page:

Limits in the form $\lim_{x\rightarrow a} f(x) - g(x)$, that is as $x \to a$, $f(x) \to \pm \infty$ and $g(x) \to \pm \infty$, then we have a limit of Indeterminate Form of Type $\infty - \infty$.

To solve these limits, we can rewrite the limit over a common denominator and solve it as an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Now let's try some examples:

## Example 1

Evaluate the following limit: $\lim_{x \to 0^+} ( \frac{1}{x} - \frac{1}{e^x - 1} )$.

As $x \to 0$, $\frac{1}{x} \to \infty$ and [[\frac{1}{e^x - 1} \to \infty ]], then we can evaluate this limit as an indeterminate difference. Let's rewrite this limit in terms of a common denominator: (1) \begin{align} \lim_{x \to 0^+} ( \frac{1}{x} - \frac{1}{e^x - 1} ) = \lim_{x \to 0^+} ( \frac{1}{x} \cdot \frac{e^x - 1}{e^x - 1} - \frac{1}{e^x - 1} \cdot \frac{x}{x} ) \\ = \lim_{x \to 0^+} \frac{e^x - 1 - x}{x(e^x - 1)} \\ \end{align} Now let's evaluate it by using L'Hospital's rule: (2) \begin{align} \lim_{x \to 0^+} \frac{e^x - 1 - x}{x(e^x - 1)} \\ = \lim_{x \to 0^+} \frac{e^x - 1}{(e^x - 1) + xe^x} \\ = \lim_{x \to 0^+} \frac{e^x}{e^x + xe^x + e^x} \\ = \lim_{x \to 0^+} \frac{e^x}{2e^x + xe^x} \\ = \frac{1}{2} \end{align} ## Example 2 Evaluate the following limit:\lim_{x \to \infty} ( x - \ln x )$. Recall that$x = \ln (e^x), hence let's apply this to our limit to obtain: (3) \begin{align} \lim_{x \to \infty} ( \ln(e^x) - \ln x ) \end{align} And now let's use the rule that\ln (a) - \ln (b) = \ln (\frac{a}{b})to obtain: (4) \begin{align} \lim_{x \to \infty} \ln \frac{e^x}{x} \\ = \ln ( \lim_{x \to \infty} \frac{e^x}{x} ) \end{align} Now let's letL = \lim_{x \to \infty} \frac{e^x}{x}$, and evaluate for$L$. Our solution to our original limit will be$\ln L\$. Notice that we can use L'Hospital's rule:

(5)
\begin{align} L = \lim_{x \to \infty} \frac{e^x}{x} \\ L = \lim_{x \to \infty} \frac{e^x}{1} \\ L = \infty \end{align}

It thus follows that:

(6)
\begin{align} \lim_{x \to \infty} ( x - \ln x ) = \ln(L) = \infty \end{align}