Skill: Relate limits to instantaneous rate of change

The instantaneous rate of change of a function at one point is found by taking the limit of its average rates of change over smaller and smaller intervals.

For a function $f(x)$ at $x=a$:

$$\text{Average rate of change} = \frac{f(a+h)-f(a)}{h}, \qquad h\ne 0$$

Here, $h$ is the change in the input. The instantaneous rate of change is

$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$

Although $h$ approaches $0$, we do not set $h=0$ in the fraction, because division by zero is undefined. Instead, we simplify first and then find the limit.

Example: Find the instantaneous rate of change of $f(x)=x^2$ at $x=3$.

Step 1: Write the average rate of change.

Here, $a=3$, so

$$\frac{f(3+h)-f(3)}{h}.$$

Step 2: Evaluate the function values.

Since $f(x)=x^2$,

$$f(3+h)=(3+h)^2$$

and

$$f(3)=3^2=9.$$

Therefore,

$$\frac{f(3+h)-f(3)}{h} = \frac{(3+h)^2-9}{h}.$$

Step 3: Simplify.

Expand and factor:

$$\frac{(3+h)^2-9}{h} = \frac{9+6h+h^2-9}{h} = \frac{6h+h^2}{h} = 6+h,$$

where $h\ne 0$.

Step 4: Let $h$ approach $0$.

$$f'(3)=\lim_{h\to 0}(6+h)=6.$$

So, the instantaneous rate of change of $f(x)=x^2$ at $x=3$ is

$$\boxed{6}.$$

Geometrically, $6$ is the slope of the tangent line to the graph at $x=3$. If $f$ represented a position function, then $6$ would be the object’s instantaneous velocity at that moment.

This is not an average rate over an interval of length zero. It is the value approached by average rates of change as the interval becomes smaller and smaller.