AP®MathsCollege BoardCalculus BCRevision NotesUnit 2: Differentiation Definition & Fundamental PropertiesFundamental Properties of DifferentiationFinding Limits using Derivatives

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

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 5 May 2026

Finding limits using derivatives

What is the connection between limits and derivatives?

You can find the value of some limits by using the connection to derivatives
- $\lim_{h \to 0} \frac{f (a + h) - f (a)}{h} = f' (a)$
- $\lim_{x \to a} \frac{f (x) - f (a)}{x - a} = f' (a)$
You can use the derivatives of common functions to derive the following results
- $\lim_{h \to 0} \frac{{(a + h)}^{n} - a^{n}}{h} = n \cdot a^{n - 1}$ and $\lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n \cdot a^{n - 1}$
- $\lim_{h \to 0} \frac{e^{a + h} - e^{a}}{h} = e^{a}$ and $\lim_{x \to a} \frac{e^{x} - e^{a}}{x - a} = e^{a}$
- $\lim_{h \to 0} \frac{\ln (a + h) - \ln a}{h} = \frac{1}{a}$ and $\lim_{x \to a} \frac{\ln x - \ln a}{x - a} = \frac{1}{a}$
- $\lim_{h \to 0} \frac{\sin (a + h) - \sin a}{h} = \cos a$ and $\lim_{x \to a} \frac{\sin x - \sin a}{x - a} = \cos a$
- $\lim_{h \to 0} \frac{\cos (a + h) - \cos a}{h} = - \sin a$ and $\lim_{x \to a} \frac{\cos x - \cos a}{x - a} = - \sin a$

How do I find the value of a limit using a derivative?

Identify the differentiable function $f$ thatmakes the limit look like one of the following:
- $\lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$
- $\lim_{x \to a} \frac{f (x) - f (a)}{x - a}$
Differentiate the function and substitute the value $x = a$
For example, for $\lim_{h \to 0} \frac{\sec (π + h) + 1}{h}$ , set $f (x) = \sec x$
- $f (π) = \sec π = \frac{1}{\cos π} = - 1$
  - This is in the form $\lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ with $a = π$
- $f' (x) = \sec x \tan x$
- Therefore, $\lim_{h \to 0} \frac{\sec (π + h) + 1}{h} = \sec π \tan π = 0$

Examiner Tips and Tricks

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

$\lim_{x \to 0} \frac{f (a + x) - f (a)}{x}$

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

Worked Example

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

Answer:

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

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

Let $f (x) = 2^{x}$

$\lim_{x \to 3} \frac{2^{x} - 8}{x - 3} = \lim_{x \to 3} \frac{f (x) - f (3)}{x - 3}$

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

$\lim_{x \to 3} \frac{f (x) - f (3)}{x - 3} = f' (3)$

Find the derivative of the function

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

Substitute $x = 3$

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

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

Worked Example

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

Answer:

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

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

Note that $\sqrt[3]{- 125} = - 5$

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

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

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

$\lim_{x \to 0} \frac{f (- 125 + x) - f (- 125)}{x} = f' (- 125)$

Find the derivative of the function

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

Substitute $x = - 125$

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

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

Examiner Tips and Tricks

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.

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

Finding limits using derivatives

What is the connection between limits and derivatives?

How do I find the value of a limit using a derivative?

Examiner Tips and Tricks

Worked Example

Worked Example

Examiner Tips and Tricks

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Jamie Wood

Reviewer: Dan Finlay