Question

Let $f:R\to R$ satisfying $|f\left(x\right)|\le x^2\forall x\in R$ is differentiable at $x=0,$ then $f^{\prime }\left(0\right)$ is

• A. 0
• B. 0.0
• C. 0.00

Answer 1

Answer: (A), (B), (C)

Since, $|f\left(x\right)|\le x^2,\forall x\in R\cdots \left(1\right)$
$\therefore$ At $x=0,|f\left(0\right)|\le 0$$\Rightarrow f\left(0\right)=0\cdots \left(2\right)$
Using first derivative principle $f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}\cdots \left(3\right)$
Now, $\left|\frac{f\left(h\right)}{h}\right|\le |h|\text{(From (1))}$
$\Rightarrow -|h|\le \frac{f\left(h\right)}{h}\le |h|$
$\Rightarrow \underset{h\to 0}{lim}\frac{f\left(h\right)}{h}=0\text{(using sandwich theorem)}\cdots \left(4\right)$
$\therefore$ from $\left(3\right)$ and $\left(4\right)$, we get $f^{\prime }\left(0\right)=0$