Example 1: L’Hospital’s Rule

Determine the following limit.

$\underset{x\to 0}{\mathit{lim}}\frac{x–\mathit{sin}\left(x\right)}{x^3}=$

Step 1: Confirm that your limit problem results in an indeterminate form, and state that L’Hopital’s Rule applies.

Like all limit problems, start with Option 1: Plug it in.

Here you would plug in 0 for x, and you will get an indeterminate form. In this case $\frac{0}{0}$.

$\underset{x\to 0}{\mathit{lim}}\frac{x–\mathit{sin}\left(x\right)}{x^3}=$

$\underset{x\to 0}{\mathit{lim}}\frac{\left(0\right)–\mathit{sin}\left(0\right)}{{\left(0\right)}^3}=\frac{0}{0}\mathit{ Indeterminate\; Form},\mathit{ L}‘\mathit{Hopital}‘\mathit{s\; Applies}$

Step 2: Take the derivative of the top equation, $f\left(x\right)$, and the bottom equation,$\mathit{ g}\left(x\right)$ , individually to find your $f\prime \left(x\right)$ and $g\prime \left(x\right)$.

$f\left(x\right)=x–\mathit{sin}\left(x\right)$

$f\prime \left(x\right)=1–\mathit{cos}\left(x\right)$

$g\left(x\right)=x^3$

$g\prime \left(x\right)={3x}^2$

Step 3: Plug your derivatives into your original limit statement, and try to evaluate the limit again using your standard limit options.

In this situation applying Option 1: Plug it in you will get an indeterminate form again , which means you need to apply L’Hopital’s Rule a second time.

$\underset{x\to 0}{\mathit{lim}}\frac{1–\mathit{cos}\left(x\right)}{{3x}^2}=$

$\underset{x\to 0}{\mathit{lim}}\frac{1–\mathit{cos}\left(0\right)}{{3\left(0\right)}^2}=\frac{0}{0}\mathit{ Indeterminate\; Form},\mathit{ L}‘\mathit{Hopital}‘\mathit{s\; Applies}$

Repeat Step 2: Take the derivative of the top equation, $f\prime \left(x\right)$, and the bottom equation, $g\prime \left(x\right)$, individually to find your $f\prime \prime \left(x\right)$ and $g\prime \prime \left(x\right)$.

$f\prime \left(x\right)=1–\mathit{cos}\left(x\right)$

$f\prime \prime \left(x\right)=\mathit{sin}\left(x\right)$

$g\prime \left(x\right)={3x}^2$

$g\prime \prime \left(x\right)=6x$

Repeat Step 3: Plug your derivatives into your original limit statement, and try to evaluate the limit again using your standard limit options.

In this situation applying Option 1: Plug it in you will get an indeterminate form again , which means you need to apply L’Hopital’s Rule a third time (I know).

$\underset{x\to 0}{\mathit{lim}}\frac{\mathit{sin}\left(x\right)}{6x}=$

$\underset{x\to 0}{\mathit{lim}}\frac{\mathit{sin}\left(0\right)}{6\left(0\right)}=\frac{0}{0}\mathit{ Indeterminate\; Form},\mathit{ L}‘\mathit{Hopital}‘\mathit{s\; Applies}$

Repeat Step 2: Take the derivative of the top equation, $f\prime \prime \left(x\right)$, and the bottom equation, $g\prime \prime \left(x\right)$, individually to find your $f\prime \prime \prime \left(x\right)$ and $g\prime \prime \prime \left(x\right)$.

$f\prime \prime \left(x\right)=\mathit{sin}\left(x\right)$

$f\prime \prime \prime \left(x\right)=\mathit{cos}\left(x\right)$

$g\prime \prime \left(x\right)=6x$

$g\prime \prime \prime \left(x\right)=6$

Repeat Step 3: Plug your derivatives into your original limit statement, and try to evaluate the limit again using your standard limit options.

And finally, after going through the L’Hopital’s Rule process three times you have a result that is not an indeterminate form.

$\underset{x\to 0}{\mathit{lim}}\frac{\mathit{cos}\left(x\right)}{6}=$

$\underset{x\to 0}{\mathit{lim}}\frac{\mathit{cos}\left(0\right)}{6}=\frac{1}{6}$

Conclusion: Due to L’Hopital’s Rule we know our original limit we were being asked to find is equal to the limit we just found in the previous step.

$\underset{x\to 0}{\mathit{lim}}\frac{x–\mathit{sin}\left(x\right)}{x^3}=\frac{1}{6}$

$\underset{x\to 0}{\mathit{lim}}\frac{x–\mathit{sin}\left(x\right)}{x^3}=\underset{x\to 0}{\mathit{lim}}\frac{1–\mathit{cos}\left(x\right)}{{3x}^2}=\underset{x\to 0}{\mathit{lim}}\frac{\mathit{sin}\left(x\right)}{6x}=\underset{x\to 0}{\mathit{lim}}\frac{\mathit{cos}\left(x\right)}{6}=\frac{1}{6}$

$\underset{x\to 0}{\mathit{lim}}\frac{x–\mathit{sin}\left(x\right)}{x^3}=\frac{1}{6}$

Final Result:

$\underset{x\to 0}{\mathit{lim}}\frac{x–\mathit{sin}\left(x\right)}{x^3}=\underset{x\to 0}{\mathit{lim}}\frac{1–\mathit{cos}\left(x\right)}{{3x}^2}=\underset{x\to 0}{\mathit{lim}}\frac{\mathit{sin}\left(x\right)}{6x}=\underset{x\to 0}{\mathit{lim}}\frac{\mathit{cos}\left(x\right)}{6}=\frac{1}{6}$

$\underset{x\to 0}{\mathit{lim}}\frac{x–\mathit{sin}\left(x\right)}{x^3}=\frac{1}{6}$

Meaning:

The overall limit a $x$ s approaches $0$ of $\frac{x–\mathit{sin}\left(x\right)}{x^3}$is $y=\frac{1}{6}$.