Squeeze Theorem:- $g(x) \leq f(x) \leq h(x)$ for all $x$ close to “a” but not necessarily at “a”,
if $\lim_{x\to a}g(x) = \lim_{x\to a }h(x)= L $ then $\lim_{x->a} f(x) =L$
Before answersing this question let’s make a note of absolute value function and signum function,
Let $h(x) = |x|$ as $x$ approaches to $0$ , $h(x)$ approaches to 0 ( $\lim_{x\to 0}|x| = 0$)
Signum function = $\dfrac{x}{|x|}$ or $\dfrac{|x|}{x}$ = $\left\{\begin{matrix} 1, & x>0\\ -1,&x<0 \end{matrix}\right.$
(C1) $| f(x) | \leq |x|$ for all real $x$
$=-| x | \leq f(x) \leq |x|$, and $-|x| $ is just constant multiple of $h(x)$.
As $\lim_{x\to 0}|x| = 0$ and $\lim_{x\to 0}-|x| = 0$ by Squeeze theorem $\lim_{x\to 0}f(x) = 0 = f(0) $
so f(x) is continuous at x=0.
(C2) $| f(x) | \leq |x|^2$ for all real $x$
$=-| x |^2 \leq f(x) \leq |x|^2$, and $-|x|^2 $ is just constant multiple of $|x|^2$.
As $\lim_{x\to 0}|x|^2 = 0$ and $\lim_{x\to 0}-|x|^2 = 0$ by Squeeze theorem $\lim_{x\to 0}f(x) = 0 = f(0) $
so f(x) is continuous at x=0.
(C2) $| f(x) | \leq |x|^3$ for all real $x$
$=-| x |^3 \leq f(x) \leq |x|^3$, and $-|x|^3 $ is just constant multiple of $|x|^3$.
As $\lim_{x\to 0}|x|^3 = 0$ and $\lim_{x\to 0}-|x|^3 = 0$ by Squeeze theorem $\lim_{x\to 0}f(x) = 0 = f(0) $
so f(x) is continuous at x=0.
Reason why i was checking for continuity of $f(x)$ at $x=0$ is that if a for given condition $f(x)$ fails to be continuous at $x=0$
then it can’t be differentiable at $x=0$, and for given all three condition $f(x)$ is continuous at $x=0$ so we can’t eliminate any option.
Differentiability:-
Function $f(x)$ is differentiable at $x=0$ if $\lim_{h\to 0 }\left[\frac{f(0+h)-f(0)}{h}\right]$ exists.
As $f(0)=0$ we can write it as $\lim_{h\to 0 }\left[\frac{f(h)}{h}\right]$, so my idea is, for a given condition if i somehow find the limit value of $\frac{f(h)}{h}$ as $h\to 0$, so i can conclude that derivative of $f(x)$ exists.
(C1) $| f(x) | \leq |x|$ as it is true for every real x, so for any arbitary h also it will be true ($h \in R$)
$| f(h) | \leq |h|$
$=-| h | \leq f(x) \leq |h|$
Now divide with h, now there will be two cases , one is h is negative and other h is positive,
Case 1:- h is postive $(h>0)$
$=\dfrac{-| h |}{h} \leq \dfrac{f(x)}{h} \leq \dfrac{|h|}{h}$
$=-1 \leq \dfrac{f(x)}{h} \leq 1$
Case2:- h is negative ($h<0$)
$=\dfrac{-| h |}{h} \geq \dfrac{f(x)}{h} \geq \dfrac{|h|}{h}$
$=1 \geq \dfrac{f(x)}{h} \geq -1$
Based on this result do you think can we conclude any thing about $\lim_{h \to 0}\dfrac{f(h)}{h}$, no because even squeeze theorem also not applicable here,
because $\lim_{h\to 0 }1 \neq \lim_{h\to 0 } -1$, even by just looking at the inequality we can say that its possible that $f’(0)$ can be any number in $[-1,1]$ and its
also possible that $f’(x)$ at $x=0$ just osillates from -1 to 1
for example consider$f(x) = x$,
$\to$ its obvious to say that $|f(x)| \leq |x|$ and we all know that derivate of $f(x) =1 $ which means $f’(0) = 1$ ,
consider $f(x) =\left\{\begin{matrix} xsin(\dfrac{1}{x}), & x\neq 0\\ 0,&x=0 \end{matrix}\right.$
$\to$ $|f(x)| \leq |x| $ its continuous at $x=0$ but its not differentiable at $x=0$( verify it.)
So by Condition C1 we can't conclude anything about differentiabilty of f(x) at $x=0$
(C2) $| f(x) | \leq |x|^2$ as it is true for every real x, so for any arbitary h also it will be true ($h \in R$)
$| f(h) | \leq |h|^2$
$=-| h |^2 \leq f(h) \leq |h|^2$
Now divide with h, now there will be two h is negative and h is positive,
Case1:- h is postive $(h>0)$
$=\dfrac{-| h |^2}{h} \leq \dfrac{f(h)}{h} \leq \dfrac{|h|^2}{h}$
Don't forget the fact that $|h|^2 = |h||h|$,
$=\dfrac{- |h||h|}{h} \leq \dfrac{f(h)}{h} \leq \dfrac{|h||h|}{h}$
$=- |h | \leq \dfrac{f(h)}{h} \leq |h|$
Case 2:- h is negative $(h<0)$
$=\dfrac{-| h |^2}{h} \geq \dfrac{f(h)}{h} \geq \dfrac{|h|^2}{h}$
$=\dfrac{-|h||h|}{h} \geq \dfrac{f(h)}{h} \geq \dfrac{|h||h|}{h}$
$= |h| \geq \dfrac{f(h)}{h} \geq -|h|$
Now comes the fun part, as $\lim_{h\to 0 }| h| = 0 $ and $\lim_{h\to 0 }-|h| =0 $ so by squeeze theorem(applicable in both cases )$\lim_{h\to 0}\dfrac{f(h)}{h} =0$,
which means f(x) is differentiable at x=0 and the value is $f’(0) = 0$ , So by having Condition 2 we can conclude that $f(x)$ is differentiable at x=0.
(C3) $| f(x) | \leq |x|^3$ as it is true for every real x, so for any arbitary h also it will be true ($h \in R$)
$| f(h) | \leq |h|^3$
$=-| h |^3 \leq f(h) \leq |h|^3$
Now divide with h, now there will be two h is negative and h is positive,
Case 1:- h is postive $(h>0)$
$=\dfrac{-| h |^3}{h} \leq \dfrac{f(h)}{h} \leq \dfrac{|h|^3}{h}$
Don't forget the fact that $|h|^3 = |h^3| = |h . h^2 | = |h|.|h^2| = |h| h^2$,
$=\dfrac{- |h|h ^2}{h} \leq \dfrac{f(h)}{h} \leq \dfrac{|h| h^2}{h}$
$=- h^2 \leq \dfrac{f(h)}{h} \leq h^2$
Case 2:- h is negative ($h<0$)
$=\dfrac{-| h |^3}{h} \geq \dfrac{f(h)}{h} \geq \dfrac{|h|^3}{h}$
$=\dfrac{- |h| h ^2}{h} \geq \dfrac{f(h)}{h} \geq \dfrac{|h| h^2}{h}$
$= h^2 \geq \dfrac{f(h)}{h} \geq -h^2$
As $\lim_{h\to 0 } h^2 = 0 $ and $\lim_{h\to 0 }-h^2 =0 $ so by squeeze theorem(applicable in both cases) $\lim_{h\to 0}\dfrac{f(h)}{h} =0$, which means f(x) is differentiable at x=0 and the value is $f’(0) = 0$, So by having Condition 3 we can conclude that $f(x)$ is differentiable at x=0.
So By condition 2 and condition 3 only we can conclude that $f(x)$ is differentiable at $x=0$, Option E will be the correct answer.
