Finding Limits using Derivatives (College Board AP® Calculus AB): Revision Note

Sometimes, the variable in the first limit is $x$ instead of $h$

$\underset{x\to 0}{\lim }\frac{f\left(a+x\right)-f\left(a\right)}{x}$

Do not get confused by this. The variable in the limit is a dummy variable, and any letter can be used.

Find $\underset{x\to 3}{\lim }\frac{2^x-8}{x-3}$.

Answer:

Substituting $x=3$ into the expression leads to $\frac{0}{0}$

Identify the function that makes the limit in the form $\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$

Let $f\left(x\right)=2^x$

$\underset{x\to 3}{\lim }\frac{2^x-8}{x-3}=\underset{x\to 3}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-3}$

$f\left(x\right)=2^x$ is differentiable, so you can use the connection between this limit and the derivative

$\underset{x\to 3}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-3}=f\text{'}\left(3\right)$

Find the derivative of the function

$f\text{'}\left(x\right)=2^x\ln 2$

Substitute $x=3$

$\begin{array}{rcl}f\text{'}\left(3\right)& =& 2^3\ln 2\\ & =& 8\ln 2\end{array}$

$\underset{x\to 3}{\lim }\frac{2^x-8}{x-3}=8\ln 2$

Find $\underset{x\to 0}{\lim }\frac{\sqrt[3]{x-125}+5}{x}$.

Answer:

Substituting $x=0$ into the expression leads to $\frac{0}{0}$

Identify the function that makes the limit in the form $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$

• Note that $\sqrt[3]{-125}=-5$

Let $f\left(x\right)=\sqrt[3]{x}$

$\underset{x\to 0}{\lim }\frac{\sqrt[3]{x-125}+5}{x}=\underset{x\to 0}{\lim }\frac{f\left(-125+x\right)-f\left(-125\right)}{x}$

$f\left(x\right)=\sqrt[3]{x}$ is differentiable at $x=-125$, so you can use the connection between this limit and the derivative

$\underset{x\to 0}{\lim }\frac{f\left(-125+x\right)-f\left(-125\right)}{x}=f\text{'}\left(-125\right)$

Find the derivative of the function

$f\text{'}\left(x\right)=\frac{1}{3}{x}^{-\frac{2}{3}}$

Substitute $x=-125$

$\begin{array}{rcl}f\text{'}\left(-125\right)& =& \frac{1}{3}·{\left(-125\right)}^{-\frac{2}{3}}\\ & =& \frac{1}{3}·\frac{1}{{\left(\sqrt[3]{-125}\right)}^2}\\ & =& \frac{1}{3}·\frac{1}{25}\\ & =& \frac{1}{75}\end{array}$

$\underset{x\to 0}{\lim }\frac{\sqrt[3]{x-125}+5}{x}=\frac{1}{75}$

In Unit 4, you learn about l'Hospital's rule which can also be used when evaluating a limit analytically, leading to $\frac{0}{0}$. If you can't identify the function, you may be able to use L'Hospital's rule instead.