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 differentiable
B) Continuous as well as differentiable
C) Neither continuous nor differentiable
D) Differentiable but not continuous

Answer 1

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.

Complete step-by-step answer:
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 \[ - c \leqslant g\left( x \right) \leqslant c\]