Proof of Various Derivative Properties — Question 6

Question 6 - Solution

By the definition of the derivative, $$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$

We are given that $$f'(a)=0.$$

Thus, $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=0.$$

To make this explicit using $\varepsilon$–$\delta$ language, let $\varepsilon>0$ be given.

Since the limit defining $f'(a)$ exists and equals $0$, there exists $\delta>0$ such that whenever $$0<|x-a|<\delta,$$ we have $$\left|\frac{f(x)-f(a)}{x-a}-0\right|<\varepsilon.$$

That is, $$\left|\frac{f(x)-f(a)}{x-a}\right|<\varepsilon.$$

This proves that $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=0.$$