# Prove that if $|f(x)| \leq x^2$, then the function is continuous and differentiable at $x=0$.

Let $f:\mathbb{R} \to \mathbb{R}$ be a function such that $|f(x)| \leq x^2$ . Prove whether or not the function is continuous and differentiable at $x=0$.

Please tell me where am i wrong i have used the sandwich theorem :

$-x^2 \leq f(x) \leq x^2$, so $\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 = \lim_{x \to 0}(-x^2) = 0$

also , $f(0)=0$

hence the function is continuous and differentiable

• this proves continuity not differentiability. – Rayees Ahmad Mar 31 '16 at 19:26

You've proved continuity and that $f(0) = 0$. To prove differentiability, you need to prove that $$\lim_{x\to 0} \frac{f(x) - f(0)}{x-0}$$ exists. However, this simplifies to proving $$\lim_{x\to 0} \frac{f(x)}{x}$$ exists. And by the given condition, we see $$\left | \frac{f(x)}{x} \right| \le \lvert x \rvert .$$ Thus $$\lim_{x\to 0} \frac{f(x)}{x} = 0$$ which shows that $f$ is differentiable at $0$ with $f'(0)=0$.
• I think you meant $|f(x)/x|\le|x|$. Also, your first limit should be $x\to0$ not $x\to\infty$. – ForgotALot Mar 31 '16 at 20:04
• but I think we should assume that $x \neq 0$ ..... correct? – Idonotknow Nov 1 '18 at 7:06