Question

# Let $f:R \to R{\text{ }}$ be a function such that $|f(x)| \leqslant {x^2}$ , for all $x \in R$. Then, at $x = 0$, f is:A) Continuous but not differentiableB) Continuous as well as differentiableC) Neither continuous nor differentiableD) Differentiable but not continuous

Hint: Here first of all we will check whether the given function f(x) is differentiable or not by using the definition of differentiability of function i.e.
A function is differentiable at point x if the following limit exists:
$\mathop {lim}\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$
Then if the function turns out to be differentiable then it would be continuous also as every differentiable function is continuous.

The given condition is:-
$|f(x)| \leqslant {x^2}$
Let us find the value of the given function at $x = 0$
Hence at $x = 0$ we get:-
$\left| {f\left( 0 \right)} \right| \leqslant {\left( 0 \right)^2} \\ \Rightarrow \left| {f\left( 0 \right)} \right| \leqslant 0 \\$
This implies:-
$f\left( 0 \right) = 0$ at $x = 0$……………………………………..(1)
Now let us check whether the f(x) is differentiable or not.
According to the definition of differentiability
A function is differentiable at point x if the following limit exists:
$\mathop {lim}\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

Hence we will check whether the given function f(x) is differentiable at $x = 0$ or not.
Hence applying the definition we get:-
$\mathop {lim}\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}$
Solving it further we get:-
$\mathop {lim}\limits_{h \to 0} \dfrac{{f(h) - f(0)}}{h}$
Now putting the value from equation 1 we get:-
$\mathop {lim}\limits_{h \to 0} \dfrac{{f(h) - 0}}{h}$
Simplifying it we get:-
$\mathop {lim}\limits_{h \to 0} \dfrac{{f(h)}}{h}$
Now it is given that:-
$|f(x)| \leqslant {x^2}$
Hence substituting h in place of x we get:-
$\left| {f\left( h \right)} \right| \leqslant {h^2}$
Dividing both the sides with h we get:-
$\left| {\dfrac{{f\left( h \right)}}{h}} \right| \leqslant \dfrac{{{h^2}}}{h} \\ \Rightarrow \left| {\dfrac{{f\left( h \right)}}{h}} \right| \leqslant h \\$
This implies:-
$- h \leqslant \dfrac{{f\left( h \right)}}{h} \leqslant h$
Now applying the limit we get:-
$\mathop {lim}\limits_{h \to 0} \left( { - h} \right) \leqslant \mathop {lim}\limits_{h \to 0} \dfrac{{f\left( h \right)}}{h} \leqslant \mathop {lim}\limits_{h \to 0} h$…………………………..(2)
Now we will evaluate the left hand limit first:-
The left hand limit is:-
$\mathop {lim}\limits_{h \to 0} \left( { - h} \right)$
Putting in the limit we get:-
$\mathop {lim}\limits_{h \to 0} \left( { - h} \right) = 0$…………………….(3)
Similarly now we will evaluate the right hand limit.
The right hand limit is given by:-
$\mathop {lim}\limits_{h \to 0} h$
Putting in the limit we get:-
$\mathop {lim}\limits_{h \to 0} h = 0$………………………………(4)
Putting the value of equation 3 and equation 4 in equation2 we get:-
$0 \leqslant \mathop {lim}\limits_{h \to 0} \dfrac{{f\left( h \right)}}{h} \leqslant 0$
Now according to sandwich theorem which states that if there exist a function $g\left( x \right)$ such that $h\left( x \right) \leqslant g\left( x \right) \leqslant f\left( x \right)$ and $\mathop {lim}\limits_{x \to 0} h\left( x \right) = \mathop {lim}\limits_{x \to 0} f\left( x \right) = c$ then $\mathop {lim}\limits_{x \to 0} g\left( x \right) = c$.
Hence applying sandwich theorem we get:-
$\mathop {lim}\limits_{h \to 0} \dfrac{{f\left( h \right)}}{h} = 0$
Therefore the limit of the given function exist at $x = 0$
Therefore the function is differentiable.
Now we know that every differentiable function is continuous.
Therefore the given function is continuous also.

Hence option B is the correct option.

Note: Students should note that every differentiable function is continuous but the converse is not true i.e. every continuous function is not differentiable.
Also, whenever a function g(x) is such that:-
$\left| {g\left( x \right)} \right| \leqslant c$
Then g(x) lies in the interval $- c \leqslant g\left( x \right) \leqslant c$