Let f: R → R be a function such that |f(x)| ≤ x ^ 2 , for all x ∈ R...

We know that if $y \le {k}^{2}, ⇒ - k \le y \le k, k \in R .$

Given \left|f\left(x\right)\right| \le {x}^{2}, \forall x \in R$</p><p>$⇒ - {x}^{2} \le f\left(x\right) \le {x}^{2} \forall x \in R$</p><p>Now,$\lim_{x \to 0}{\left(- x\right)}^{2} = \lim_{x \to 0}\left({x}^{2}\right) = 0,$</p><p>Hence, by using sandwich theorem,$\lim_{x \to 0}f\left(x\right) = 0$</p><p>Further \left|f\left(0\right)\right| \le 0 ⇒ f\left(0\right) = 0$

Since, $\lim_{x \to 0}f\left(x\right) = f\left(0\right) = 0$

Hence, $f\left(x\right)$ is continuous at $x = 0$

Now, if $x \ne 0,$ then $- \left(x\right) \le \dfrac{f\left(x\right)}{\left(x\right)} \le \left(x\right)$

Thus, by using sandwich theorem, we have $\lim_{x \to 0} - \left(x\right) \le \lim_{x \to 0}\dfrac{f\left(x\right)}{\left(x\right)} \le \lim_{x \to 0}\left(x\right)$

$⇒ 0 \le \lim_{x \to 0}\dfrac{f\left(x\right)}{\left(x\right)} \le 0$

$⇒ \lim_{x \to 0}\dfrac{f\left(x\right)}{\left(x\right)} = 0 . . . \left(i\right)$

We know that a function $f\left(x\right)$ is differentiable at a point $x = a,$ if $\lim_{h \to 0}\dfrac{f\left(a - h\right) - \left(a\right)}{- h} = \lim_{h \to 0}\dfrac{f\left(a + h\right) - \left(a\right)}{h}$

Now, the left-hand derivative at $x = 0$ is

$\lim_{h \to 0}\dfrac{f\left(0 - h\right) - f\left(0\right)}{- h} = \lim_{h \to 0}\dfrac{f\left(- h\right)}{- h} = 0,$ ( by using equation \left(i\right)$) </p><p>And, the right-hand derivative at$x = 0$is</p><p>$\lim_{h \to 0}\dfrac{f\left(0 + h\right) - f\left(0\right)}{h} = \lim_{h \to 0}\dfrac{f\left(h\right)}{h} = 0,$( by using equation \left(i\right)$)

Hence, $f\left(x\right)$ is differentiable at $x = 0 .$