Boundedness of third derivative and limit of quotient

Question

I came across the following problem:

Assuming that $f\colon \mathbb{R}\to\mathbb{R}$ is twice differentiable at $x_0$, define $$ E(h)=f''(x_0)-\frac{1}{h^2}\left[ f(x_0+h)-2f(x_0)+f(x_0-h). \right] $$ Suppose $f^{(3)}$ (the third derivative of $f$) is bounded in $\mathbb{R}$. Prove or disprove that $\lim_{h\to 0^+}(1/h)\cdot E(h)=0$.

I was trying to show that this is true. I used L'Hôpital's Rule and the limit turned out to be zero. But I'm struggling to see why we need boundedness of the third derivative. Could somebody give me a hint on this?

I think we just need $f'''$ to exist for every $x\in \mathbb{R}$, since that is enough to guarantee the differentiability of $f''$ at every point. I tried using unbounded functions like $f'''(x)=e^x$ and $f'''(x)=x^3$, and the limit is still zero.

You need to find a function whose third derivative is unbounded at $x_0$. — Martin R, Feb 08 '25 at 20:35
@MartinR doesn't the definition of third derivative at $x_0$ imply that $f'''(x_0)\in\mathbb{R}$? — P3p3O, Feb 08 '25 at 20:39
What I mean is that $f'''$ is unbounded in every neighborhood of $x_0$. — Martin R, Feb 08 '25 at 20:40
Boundedness of third derivative is unnecessary here. The result holds under the weaker hypothesis of existence of $f''' (x_0)$. Just use Taylor theorem with Peano form of remainder. L'Hospital's rule can also be used (but not more than 2 times). — Paramanand Singh, Feb 09 '25 at 03:07

Boundedness of third derivative and limit of quotient

0 Answers0