Solutions

a) Let

$$g(x)=\frac{f(x)-f(a)}{x-a}.$$

By Exercise 5-9, we have

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

b) We have

$$f(x)=g(x)\implies f(x)-f(a)=g(x)-g(a).$$

Thus,

$$\frac{f(x)-f(a)}{x-a}=\frac{g(x)-g(a)}{x-a}.$$

Since $f(x)=g(x)$ for all $x$ in an open interval containing $a$, by Exercise 5-11 and part a) of this exercise,

$$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{g(x)-g(a)}{x-a}=g'(a).$$