Shrinking the interval (h→0)

For $y=x^2$, the average rate of change over the interval of width $h$ starting at $x=a$ is $2a+h$ (for $h\neq 0$).

With $a=3$, taking $h=0.1$, $0.01$, $0.001$ gives $6.1$, $6.01$, $6.001$, approaching $6=2a$. Approaching from the negative side gives $5.9$, $5.99$, $5.999$, which approaches the same $6$.

This limiting value is the slope of the tangent at $x=a$. Later it is defined as the derivative $f'(a)$.

Examples

1. 01Average rate of change of $y=x^2$ at $x=a$. As the width $h$ approaches $0$, what value does it approach?$\displaystyle 2a+h$AnswerHide

   Answer$\displaystyle 2a$
2. 02Average rate of change of $y=x^2$ at $x=5$. As the width $h$ approaches $0$, what value does it approach?$\displaystyle \begin{matrix} h=-0.1 & \Rightarrow & 9.9 \\ h=-0.01 & \Rightarrow & 9.99 \\ h=-0.001 & \Rightarrow & 9.999 \end{matrix}$AnswerHide

   Answer$\displaystyle 10$
3. 03For which h is the average rate of change from $x=1$ to $x=1+h$ closest to the slope of the tangent?$\displaystyle y=x^2$AnswerHide

   Answer$\displaystyle h=0.001$
4. 04Average rate of change of $y=x^2$ at $x=3$. As the width $h$ approaches $0$, what value does it approach?$\displaystyle \begin{matrix} h=-0.1 & \Rightarrow & 5.9 \\ h=-0.01 & \Rightarrow & 5.99 \\ h=-0.001 & \Rightarrow & 5.999 \end{matrix}$AnswerHide

   Answer$\displaystyle 6$
5. 05Average rate of change of $y=x^2$ at $x=4$. As the width $h$ approaches $0$, what value does it approach?$\displaystyle \begin{matrix} h=-0.1 & \Rightarrow & 7.9 \\ h=-0.01 & \Rightarrow & 7.99 \\ h=-0.001 & \Rightarrow & 7.999 \end{matrix}$AnswerHide

   Answer$\displaystyle 8$
6. 06Average rate of change of $y=x^2$ at $x=1$. As the width $h$ approaches $0$, what value does it approach?$\displaystyle \begin{matrix} h=-0.1 & \Rightarrow & 1.9 \\ h=-0.01 & \Rightarrow & 1.99 \\ h=-0.001 & \Rightarrow & 1.999 \end{matrix}$AnswerHide

   Answer$\displaystyle 2$

FAQ

Q1Why not just put $h=0$ from the start?AnswerHide

A

Putting $h=0$ into the original $\dfrac{(a+h)^2-a^2}{h}$ gives $\dfrac{0}{0}$, which has no value. Keep $h\neq 0$, cancel to get $2a+h$, and then let $h$ approach $0$.

Q2Does it approach the same value when $h$ is negative?AnswerHide

A

For $y=x^2$, yes. $2a+h$ approaches $2a$ as $h\to 0$ whether $h$ is positive or negative, so shrinking from either side gives the same value.