Question

Instead of choosing small values of h, we could have chosen values of z close to c. What limit involving z instead of h is equivalent to the one involving h?

Short answer

The derivative value is$f^{\text{'}}\left(c\right)=\underset{z\to c}{\lim }\frac{f\left(z\right)-f\left(c\right)}{z-c}.$

Step-by-step solution

Step 1.Given Information

Given as h tends to zero, z close to c

Step 2.Forming limits 

If the limit of these quantities approaches a real number as $h\to 0$, or as $z\to c$, then we

will define that real number to be the derivative of f at the point $x=c$.

Step 3.The solution 

As we know that $z=c+h$as $h\to 0;z\to c.$The value of $h=z-c.$

The derivative value is$$\begin{gathered} f\text{'}\left(c\right)=\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h} \\ {f}^{\text{'}}\left(c\right)=\underset{z\to c}{\lim }\frac{f\left(z\right)-f\left(c\right)}{z-c}. \end{gathered}$$