Proof using Taylor expansion

Question

Let $f:(a,b) \rightarrow \mathbb{R}$ be twice differentiable at $x_0$. Assume $$\lim_{x\rightarrow 0} \frac{f(x_0+2x)-f(x_0+x)}{x^3}=0$$ Prove $f'(x_0)=f''(x_0)=0$.

A hint suggests using a Taylor expansion of $f$, but I am unsure of how to do it.

Answer 1

We know that in general $g(y)=g(a)+g'(a)(y-a)+\frac{g''(a)}{2!}(y-a)^2+\mathcal{O}(y^3)$, so substitute $a=x_0$ and $y=x_0+2x$ for the first, $y=x_0+x$ for the other, to obtain:

$f(x_0+2x)=f(x_0)+2f'(x_0)x+2f''(x_0)x^2+\mathcal{O}(x^3)$

and

$f(x_0+x)=f(x_0)+f'(x_0)x+\frac{f''(x_0)}{2}x^2+\mathcal{O}(x^3)$

Now we get

$$0=\lim_{x\rightarrow 0} \frac{f(x_0+2x)-f(x_0+x)}{x^3}=$$ $$\lim_{x\rightarrow 0} \frac{f(x_0)+2f'(x_0)x+2f''(x_0)x^2-f(x_0)-f'(x_0)x-\frac{f''(x_0)}{2}x^2+\mathcal{O}(x^3)}{x^3}=$$ $$\lim_{x\rightarrow 0}\frac{f'(x_0)}{x^2}+\lim_{x\rightarrow 0}\frac{3f''(x_0)}{2x}+\lim_{x\rightarrow 0}\mathcal{O}(1)$$ which can only exist if $f'(x_0)=f''(x_0)=0$ (and here also $\mathcal{O}(1)=0$).

Answer 2

Hint: Taylor's Theorem says $$ f(x_0+t)=f(x_0)+f'(x_0)t+f''(x_0)\frac{t^2}2+o\!\left(t^2\right) $$