Question

If $R\to R$ be a function we sat that f has
Property 1: If \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}}\] exists and is finite.
Property 2: If \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{{{h}^{2}}}\] exists and is finite.
Then, which of the following options is/are correct.
\[\begin{align}
  & A.f\left( x \right)={{x}^{\dfrac{2}{3}}}\text{ has property 1} \\
 & \text{B}\text{.f}\left( x \right)=\sin x\text{ has property 2} \\
 & \text{C}\text{.f}\left( x \right)=\left| x \right|\text{ has property 1} \\
 & \text{D}\text{.f}\left( x \right)=x\left| x \right|\text{ has property 2} \\
\end{align}\]

Answer 1

Hint: To solve this question, we will consider all options separately and check which one of them satisfy property 1 or 2. Also, in mid of solution we will use the fact that, if $\displaystyle \lim_{h \to 0},\dfrac{{{h}^{m}}}{{{h}^{n}}}\text{ and m n}$ then $\displaystyle \lim_{h \to 0},\dfrac{{{h}^{m}}}{{{h}^{n}}}=0$ and if $\text{m }<\text{ n}$ then $\displaystyle \lim_{h \to 0},\dfrac{{{h}^{m}}}{{{h}^{n}}}\text{ }=\text{ }\dfrac{1}{0}\text{ }=\text{ }\infty $ does not exist.
While calculating options having sin x we will use formula of limit with sin which is given as $\displaystyle \lim_{h \to 0},\dfrac{\sin x}{x}\text{ }=1$

Complete step-by-step answer:
Given,
Property 1: If \[\displaystyle \lim_{h \to 0},\text{ }\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}}\] exists and is finite.
Property 2: If \[\displaystyle \lim_{h \to 0},\text{ }\dfrac{f\left( h \right)-f\left( 0 \right)}{{{h}^{2}}}\] exists and is finite.
To solve this question, we will consider all options one by one and see which one is correct.
Consider option A.
A. $f\left( x \right)={{x}^{\dfrac{2}{3}}}$ has property 1.
Let us compute \[\displaystyle \lim_{h \to 0},\text{ }\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}}\] for $f\left( x \right)={{x}^{\dfrac{2}{3}}}$
Substituting $f\left( h \right)={{h}^{\dfrac{2}{3}}}$ in above we get:
\[\displaystyle \lim_{h \to 0},\text{ }\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}}\] and f(0) = 0 so we have
\[\displaystyle \lim_{h \to 0},\text{ }\dfrac{{{h}^{\dfrac{2}{3}}}-0}{\sqrt{\left| h \right|}}\Rightarrow \displaystyle \lim_{h \to 0},\text{ }\dfrac{{{h}^{\dfrac{2}{3}}}}{\sqrt{\left| h \right|}}\]
Now,
\[\sqrt{h}={{h}^{\dfrac{1}{2}}}\Rightarrow \displaystyle \lim_{h \to 0},\text{ }\dfrac{{{h}^{\dfrac{2}{3}}}}{{{\left| h \right|}^{\dfrac{1}{2}}}}\]
Now because $\dfrac{2}{3}\text{ }>\text{ }\dfrac{1}{2}$
So, power of numerator is bigger than power of denominator and as $h\text{ }\to 0$ so whether $h\text{ }\to {{0}^{+}}\Rightarrow h\text{ }\to {{0}^{-}}$ the numerator of $\dfrac{{{h}^{\dfrac{2}{3}}}}{{{\left| h \right|}^{\dfrac{1}{2}}}}$ is always greater than its denominator.
\[\Rightarrow \displaystyle \lim_{h \to 0},\text{ }\dfrac{{{h}^{\dfrac{2}{3}}}}{\sqrt{\left| h \right|}}=0\] as we have a numerator bigger power.
So \[\displaystyle \lim_{h \to 0},\text{ }\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}}\] exist and is finite for $f\left( x \right)={{x}^{\dfrac{2}{3}}}$
So, option A is correct.
A. $f\left( x \right)={{x}^{\dfrac{2}{3}}}$ has property 1.
Let us compute \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}}\] for $f\left( x \right)={{x}^{\dfrac{2}{3}}}$
Substituting $f\left( h \right)={{h}^{\dfrac{2}{3}}}$ in above we get:
\[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}}\] and f(0) = 0 so we have
\[\begin{align}
  & \displaystyle \lim_{h \to 0},\dfrac{{{h}^{\dfrac{2}{3}}}-0}{\sqrt{\left| h \right|}} \\
 & \displaystyle \lim_{h \to 0},\dfrac{{{h}^{\dfrac{2}{3}}}}{\sqrt{\left| h \right|}} \\
\end{align}\]
Now,
\[\begin{align}
  & \sqrt{h}={{h}^{\dfrac{1}{2}}} \\
 & \displaystyle \lim_{h \to 0},\dfrac{{{h}^{\dfrac{2}{3}}}}{{{\left( \left| h \right| \right)}^{\dfrac{1}{2}}}} \\
\end{align}\]
Now because $\dfrac{2}{3}>\dfrac{1}{2}$
So, power of numerator is bigger than power of denominator and as $h\to 0$ so whether $h\to {{0}^{+}}\Rightarrow h\to {{0}^{-}}$ the numerator of $\dfrac{{{h}^{\dfrac{2}{3}}}}{{{\left( \left| h \right| \right)}^{\dfrac{1}{2}}}}$ is always greater than its denominator.
\[\Rightarrow \displaystyle \lim_{h \to 0},\dfrac{{{h}^{\dfrac{2}{3}}}}{\sqrt{\left| h \right|}}=0\] as we have a numerator bigger power.
So \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}}\] exist and is finite for $f\left( x \right)={{x}^{\dfrac{2}{3}}}$
So, option A is correct.
Consider option B.
B. f(x) = sin x has property 2.
We have, property 2 has \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{{{h}^{2}}}\]
Substituting f(h) = sin h in above and f(0) = sin 0 =0 we get:
\[\begin{align}
  & \displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{{{h}^{2}}} \\
 & \displaystyle \lim_{h \to 0},\dfrac{\left( \sinh -\sin 0 \right)}{{{h}^{2}}} \\
 & \displaystyle \lim_{h \to 0},\dfrac{\left( \sinh -0 \right)}{{{h}^{2}}} \\
\end{align}\]
Taking $\dfrac{1}{h}$ common we get:
\[\displaystyle \lim_{h \to 0},\dfrac{1}{h}\left( \dfrac{\sinh }{h} \right)\]
Now, we will apply property of product of limit which says $\lim \left( ab \right)=\lim a\cdot \lim b$
\[\begin{align}
  & \displaystyle \lim_{h \to 0},\dfrac{1}{h}\left( \dfrac{\sinh }{h} \right) \\
 & \Rightarrow \left( \displaystyle \lim_{h \to 0},\dfrac{1}{h} \right)\left( \displaystyle \lim_{h \to 0},\dfrac{\sinh }{h} \right) \\
\end{align}\]v
Again, we will use property of limit of sin which says:
\[\begin{align}
  & \displaystyle \lim_{h \to 0},\dfrac{\sin x}{x}=1 \\
 & \Rightarrow \left( \displaystyle \lim_{h \to 0},\dfrac{1}{h} \right)\left( \displaystyle \lim_{h \to 0},\dfrac{\sinh }{h} \right)=\left( \displaystyle \lim_{h \to 0},\dfrac{1}{h} \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
 & \Rightarrow \left( \displaystyle \lim_{h \to 0},\dfrac{1}{h} \right)=\dfrac{1}{0}=\infty \\
\end{align}\]
So, this limit doesn't exist.
Hence, option B is wrong.
Consider option C.
C. $f\left( x \right)=\left| x \right|$ has property 1.
Consider \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}}\] and substituting $f\left( x \right)=\left| x \right|$ in above we get:
\[\begin{align}
  & \displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}} \\
 & \displaystyle \lim_{h \to 0},\dfrac{\left| h \right|-0}{\sqrt{\left| h \right|}} \\
\end{align}\]
Again using
\[\begin{align}
  & \sqrt{h}={{h}^{\dfrac{1}{2}}} \\
 & \displaystyle \lim_{h \to 0},\dfrac{\left| h \right|}{{{\left| h \right|}^{\dfrac{1}{2}}}} \\
\end{align}\]
Observing we have power of $\left| h \right|$ in numerator is 1 and power of $\left| h \right|$ in denominator is $\dfrac{1}{2}$ and $1>\dfrac{1}{2}$
Numerator is bigger in \[\dfrac{\left| h \right|}{{{\left| h \right|}^{\dfrac{1}{2}}}}\] than denominator.
\[\displaystyle \lim_{h \to 0},\dfrac{\left| h \right|}{{{\left| h \right|}^{\dfrac{1}{2}}}}=0\] which is a finite number.
So, \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{\sqrt{\left| h \right|}}\] exist and is finite for $f\left( x \right)=\left| x \right|$
So option C is correct.
Consider option D.
D. $f\left( x \right)=x\left| x \right|$ has property 2.
Consider \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{{{h}^{2}}}\]
Applying $f\left( x \right)=h\left| h \right|$ in above we get:
\[\displaystyle \lim_{h \to 0},\dfrac{h\left| h \right|-0}{{{h}^{2}}}\]
When $h>0\Rightarrow \displaystyle \lim_{h \to 0},\dfrac{h\left| h \right|-0}{{{h}^{2}}}=\displaystyle \lim_{h \to 0},\dfrac{{{h}^{2}}}{{{h}^{2}}}=1$
And when $h<0\Rightarrow \displaystyle \lim_{h \to 0},\dfrac{h\left| h \right|-0}{{{h}^{2}}}=\displaystyle \lim_{h \to 0},\dfrac{-{{h}^{2}}}{{{h}^{2}}}=-1$
This is so because $\left| x \right|=x$ when $x>0$ and $\left| x \right|=-x$ when $x<0$
So, when $h\to {{0}^{+}}$ then \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{{{h}^{2}}}=+1\]
And when $h\to {{0}^{-}}$ then \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{{{h}^{2}}}=-1\]
So, we have two different values, so \[\displaystyle \lim_{h \to 0},\dfrac{f\left( h \right)-f\left( 0 \right)}{{{h}^{2}}}\] doesn't exist when $f\left( x \right)=x\left| x \right|$
So, option D is wrong.

So, the correct answers are “Option A and C”.

Note: While considering \[\displaystyle \lim_{h \to 0},\dfrac{\left| h \right|}{{{\left| h \right|}^{\dfrac{1}{2}}}}\] in option C we can also go for another method which is as below:
\[\displaystyle \lim_{h \to 0},\dfrac{\left| h \right|}{{{\left| h \right|}^{\dfrac{1}{2}}}}=\displaystyle \lim_{h \to 0},{{\left| h \right|}^{1-\dfrac{1}{2}}}\]
This is so as $\dfrac{{{a}^{m}}}{{{a}^{n}}}{{a}^{m-n}}$
\[\displaystyle \lim_{h \to 0},{{\left| h \right|}^{1-\dfrac{1}{2}}}=\displaystyle \lim_{h \to 0},{{\left| h \right|}^{\dfrac{1}{2}}}\]
So, after we apply limit \[\displaystyle \lim_{h \to 0},{{\left| h \right|}^{\dfrac{1}{2}}}=0\]
Similarly, in option A we have \[\displaystyle \lim_{h \to 0},\dfrac{{{h}^{\dfrac{2}{3}}}}{{{\left( \left| h \right| \right)}^{\dfrac{1}{2}}}}\]
\[\displaystyle \lim_{h \to 0},{{\left| h \right|}^{\dfrac{2}{3}-\dfrac{1}{2}}}\Rightarrow \displaystyle \lim_{h \to 0},{{\left| h \right|}^{\dfrac{1}{6}}}\]
Which is anyway 0.