Proving differentiability of $f$ from bound

Question

Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a function such that $|f(x)|\leq 2 \tan (\|x\| ^{3/2})$ in a neighborhood of $0$. Prove that $f$ is differentiable at $0$.

I don't know how to approach this problem. Any hint would be appreciated.

Answer 1

You are given $f(0) = 0$. You must show that there exists a linear map $A : \mathbb R^n \to \mathbb R$ with the property that $$f(x) - f(0) = Ax + o(\|x\|)$$ as $x \to 0$. If one exists, it is unique.

The hypothesis of the problem is pretty strong. For $x$ in a neighborhood of $0$ you have $$\frac{|f(x) - f(0)|}{\|x\|} \le \frac{2 \tan \|x\|^{3/2}}{\|x\|} = \|x\|^{1/2} \frac{2 \tan \|x\|^{3/2}}{\|x\|^{3/2}} \to 0$$ as $x \to 0$ because $\lim_{t \to 0} \dfrac{\tan t}{t} = 1$.

Thus $f(x) - f(0) = o(\|x\|)$ so you have $A = 0$.

Answer 2

$f'(0) = \lim_\limits{\|\Delta x\|\to 0}\frac {f(\Delta x)-f(0)}{\|\Delta x\|}$

$f(0) = 0$ and $|f(\Delta x)| < 2\tan \|\Delta x\|^{\frac 32}$

Substitute $h = \|\Delta x\|$

$|f'(0)| < \lim_\limits {h\to 0^+} \frac {\tan h^\frac 32}{h}$

And now show that that limit equals $0.$