Q. 5 - Calculus

Question

Explain why the limits $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ and $\underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x}$ are the same for any function $f$ . (Hint: Consider the substitution $z=x+h$.)

Step-by-Step Solution

Answer

$\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ and $\underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x}$are the same for any function $f$.

Step 1. Given information

We have to prove that $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ and $\underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x}$ are the same for any function $f$.

Step 2. Proof of the question

Substitute $z=x+h$

When $z\to x$, the value of $h\to 0$

$$\begin{gathered} \underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x} \\ =\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{x+h-x} \\ =\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \end{gathered}$$

Hence it is proved.