The derivative of a function is defined as the limit, which finds the slope of the tangent line to a function. Therefore, we can find an equation by recalling that the slope of a line is found by dividing the change in *y* by the change in *x*. This means that we can use limits to find the derivatives of functions.

Here, we will learn how to find derivatives of functions using limits. We will learn a formula and apply it to solve some practice problems.

## Formula for finding derivatives using limits

To find the derivative of a function using limits, we can use the following formula:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

This formula has the following important parts:

- $latexf(x+h)$. This means that we have to evaluate the given function using the input $latex x+h$.
- $latexf(x)$. This is the original function in terms of
*x*. *h*in the denominator. This value must remain the same until we evaluate the limit or until it is simplified.- $latex \lim _{h \to 0}$ is the limit. Generally, we solve it by substituting 0 for all the
*h*that we find.

Therefore, to apply this formula, we have to start by finding the expression $latex f(x+h)$ in the numerator. Then, we subtract the original function $latex f(x)$.

Generally, we can simplify the *h* in the denominator with the *h* in the numerator. Finally, we use $latex h=0$ to solve the limit (this will work in most cases, but not always).

## Finding derivatives using limits – Examples with answers

The formula for derivatives using limits is applied to solve the following examples. Each example has its respective solution, but try to solve the problems yourself first.

**EXAMPLE 1**

Use limits to find the derivative of $latex f(x)=5x$.

##### Solution

The formula for the derivative of a function using limits is:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Now, we use the $latex function f(x)=5x$ to rewrite the numerator. Therefore, we have:

$$f'(x)=\lim _{h \to 0}\frac{5(x+h)-5x}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{5x+5h-5x}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{5h}{h}$$

We can simplify the *h* of the numerator with the *h* of the denominator

$$f'(x)=\lim _{h \to 0}(5)$$

Since we don’t have any *h*, we simply remove the limit:

$latex f'(x)=5$

**EXAMPLE **2

**EXAMPLE**

Find the derivative of $latex f(x)=x^2$ using limits.

##### Solution

We start by writing the formula for derivatives with limits:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Now, we rewrite the numerator using $latex f(x)=x^2$. Thus, we have:

$$f'(x)=\lim _{h \to 0}\frac{(x+h)^2-x^2}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{x^2+2hx+h^2-x^2}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{2hx+h^2}{h}$$

We can factor out the *h* from the numerator and simplify with the denominator:

$$f'(x)=\lim _{h \to 0}\frac{h(2x+h)}{h}$$

$$f'(x)=\lim _{h \to 0}(2x+h)$$

Finally, we solve the limit by substituting $latex h=0$ into the expression:

$latex f'(x)=2x+0$

$latex f'(x)=2x$

**EXAMPLE **3

**EXAMPLE**

Find the derivative of $latex f(x)=2x^2+3$ using limits.

##### Solution

We have the following formula:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Then, we use the function $latex f(x)=2x^2+3$ to rewrite the numerator. Thus, we have:

$$f'(x)=\lim _{h \to 0}\frac{(2(x+h)^2+3)-(2x^2+3)}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{2(x^2+2hx+h^2)+3-2x^2-3}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{2x^2+4hx+2h^2+3-2x^2-3}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{4hx+2h^2}{h}$$

Factoring the *h* from the numerator and simplifying with the *h* from the denominator, we get:

$$f'(x)=\lim _{h \to 0}\frac{h(4x+2h)}{h}$$

$$f'(x)=\lim _{h \to 0}(4x+2h)$$

Finally, we solve the limit by substituting $latex h=0$ into the expression:

$latex f'(x)=4x+2(0)$

$latex f'(x)=4x$

**EXAMPLE **4

**EXAMPLE**

Use limits to find the derivative of $latex f(x)=3x^2+5x$.

##### Solution

We start by writing the formula to find derivatives using limits:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Now, we use the function $latex f(x)=3x^2+5x$ on the numerator, and we have:

$$f'(x)=\lim _{h \to 0}\frac{(3(x+h)^2+5(x+h))-(3x^2+5x)}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{3(x^2+2hx+h^2)+5x+5h-3x^2-5x}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{3x^2+6hx+3h^2+5x+5h-3x^2-5x}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{6hx+3h^2+5h}{h}$$

We factor the *h* of the numerator to simplify with the *h* of the denominator:

$$f'(x)=\lim _{h \to 0}\frac{h(6x+3h+5)}{h}$$

$$f'(x)=\lim _{h \to 0}(6x+3h+5)$$

Finally, we solve the limit by substituting $latex h=0$ into the expression:

$latex f'(x)=6x+3(0)+5$

$latex f'(x)=6x+5$

**EXAMPLE **5

**EXAMPLE**

Find the derivative of $latex f(x)=\frac{1}{x}$ using limits.

##### Solution

We have the formula:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Using the function $latex f(x)=\frac{1}{x}$ on the numerator, we have:

$$f'(x)=\lim _{h \to 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$$

We can use the common denominator $latex (x+h)x$ to subtract the fractions from the numerator:

$$f'(x)=\lim _{h \to 0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{\frac{x-x-h}{x^2+xh)}}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{\frac{-h}{x^2+xh)}}{h}$$

We can rewrite the fraction and simplify the *h*:

$$f'(x)=\lim _{h \to 0}\frac{-h}{h(x^2+xh)}$$

$$f'(x)=\lim _{h \to 0}\frac{-1}{x^2+xh}$$

Finally, we solve the limit by substituting $latex h=0$ into the expression:

$$ f'(x)=\frac{-1}{x^2+x(0)}$$

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

**EXAMPLE **6

**EXAMPLE**

Determine the derivative of $latex f(x)=\sqrt{x}$ using limits.

##### Solution

We have the formula:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Now, we rewrite the numerator using the function $latex f(x)=\sqrt{x}$. Therefore, we have:

$$f'(x)=\lim _{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$$

We can multiply both the numerator and the denominator by the conjugate of the numerator:

$$f'(x)=\lim _{h \to 0}\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}$$

$$f'(x)=\lim _{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$$

$$f'(x)=\lim _{h \to 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$$

We simplify the *h* in the numerator with the *h* in the denominator:

$$f'(x)=\lim _{h \to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}$$

Finally, we solve the limit by substituting $latex h=0$ into the expression:

$$f'(x)=\lim _{h \to 0}\frac{1}{\sqrt{x+0}+\sqrt{x}}$$

$$f'(x)=\lim _{h \to 0}\frac{1}{\sqrt{x}+\sqrt{x}}$$

$$f'(x)=\lim _{h \to 0}\frac{1}{2\sqrt{x}}$$

## Finding derivatives using limits – Practice problems

Use the formula for finding derivatives of functions using limits to solve the following practice problems.

## See also

Interested in learning more about derivatives of functions? You can take a look at these pages:

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