$f$ is differenciable at a point a iff $f(x)=f(a)+L(x-a)+s(x)(x-a)$, with $L=f'(a)$ and $s(x) \to 0$ when $x \to a$

Question

I already prove that $f(x)=f(a)+L(x-a)+s(x)(x-a)$ implies that $f$ differenciable in $a$. But i dont know how prove the other side.

Some hint, please.

Answer 1

If $f$ is differentiable at $x=a$ with the existence of the derivative $f'(a)$, now let $s(a)=0$ and $s(x)=\dfrac{f(x)-f(a)}{x-a}-f'(a)$ for $x\ne a$, then $s(x)(x-a)=f(x)-f(a)-f'(a)(x-a)=f(x)-f(a)-L(x-a)$ and we have $\lim_{x\rightarrow a}s(x)=\lim_{x\rightarrow a}(f(x)-f(a)-L(x-a))=f(a)-f(a)-L(a-a)=0$.