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Well the question presented to me is this. The given function is,
$$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{2}x + 2,\,\;\;\;x < 2}\\{\sqrt {2x} ,\;\;\;\;\;\;x \ge 2}\end{array}} \right. $$ Now have to check whether the given function is differenciable at $x=2$ ?


My approach:
For this function to be differentiable, the left hand derivative and the right hand derivative must exits, and both be equal.
Left hand derivative:
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ - }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left( {\frac{1}{2}x + 2} \right) - \left( {\frac{1}{2}\left( 2 \right) + 2} \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left( {\frac{1}{2}x + 2} \right) - 3}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\frac{1}{2}x - 1}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{x - 2}}{{2\left( {x - 2} \right)}}\end{array} $$
This gives,
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ - }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{}}{{2}}\\ = \frac{1}{2}\end{array} $$


Right hand derivative:
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ + }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\left( {\sqrt {2x} } \right) - \left( {\sqrt {2\left( 2 \right)} } \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {2x} - \sqrt 4 }}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {2x} - 2}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {2x} - 2}}{{x - 2}} \times \frac{{\sqrt {2x} + 2}}{{\sqrt {2x} + 2}}\end{array} $$
This gives,
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ + }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{2x - 4}}{{\left( {x - 2} \right)\left( {\sqrt {2x} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{2}{{\left( {\sqrt {2x} + 2} \right)}}\\ = \frac{2}{{2 + 2}}\\ = \frac{1}{2}\end{array} $$


Problem is right hand derivative and left hand derivative are coming same. Fooling me that its differentiable at $x=2$.


The graph of the function is,
$f(x)$ Graph
The function is discontinues at $x=2$. So, its not differentiable at $x=2$.


Where have I gone wrong ? How do I prove analytically without taking help of graph that the function is not differentiable at $x=2$?

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  • $\begingroup$ You have made a mistake in value of $f(2)$. It is given by $f(2)=2$ and not $3$ as you have tried to do while calculating left hand derivative. The left hand derivative is $-\infty$. Also note that the analytical thing is the real stuff and graphical arguments are only an approximation to the analytical argument and are meant to help in understanding the concepts. $\endgroup$ – Paramanand Singh May 25 '17 at 14:27
  • $\begingroup$ @ParamanandSingh I think it would be correct to take this as function $ \frac{1}{2}x + 2 $ when approaching from left. $\endgroup$ – Roopesh Singh May 25 '17 at 14:45
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    $\begingroup$ The value $f(2)$ does not depend on left right. It depends on point $2$. Do not confuse value of a function with left or right limit of the function. These are different concepts and value of a function is a much simpler concept than the limit of a function. $\endgroup$ – Paramanand Singh May 25 '17 at 14:45
  • $\begingroup$ Yes.. Got the point. I did a mistake there. Thanks. $\endgroup$ – Roopesh Singh May 25 '17 at 14:47
  • $\begingroup$ Do you seriously need help in getting values of $f$? Check definition of $f$. If $x<2$ the value $f(x) $ is given by formula $(x/2)+2$ so $f(1)=5/2$. But if $x\geq 2$ then $f(x) $ is given by formula $\sqrt{2x}$ and hence $f(2)=\sqrt{4}=2$. I hope this is clear. $\endgroup$ – Paramanand Singh May 25 '17 at 14:51
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$$f (2^-)=\lim_{x\to 2^-} (\frac {x}{2}+2)=1+2=3$$ $$ f (2^+)=\lim_{2^+}\sqrt {2x}=2$$

$f $ is not continuous at $x=2$ thus it is not differentiable at $x=2$.

By definition, differentiable at $x=x_0$ means $$\exists L\in \mathbb R \;\exists \eta>0 :\forall x\in (2-\eta,2+\eta) $$

$$f (x)=f (2)+(x-2)\Bigl (L+\epsilon (x)\Bigr) $$

with $$\lim_{x\to 2}\epsilon (x)=0$$

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  • $\begingroup$ Have you arrived at $$f (2^-)=1+2=3\neq f (2^+)=2$$ without using help of graph ? If yes, please do explain a bit. $\endgroup$ – Roopesh Singh May 25 '17 at 12:08
  • $\begingroup$ @Mathanatic Look now $\endgroup$ – hamam_Abdallah May 25 '17 at 12:12
  • $\begingroup$ What's wrong in what I did? Is that definition incorrect? + How $$f (2^-)=1+2=3\neq f (2^+)=2$$ came ? $\endgroup$ – Roopesh Singh May 25 '17 at 12:17
  • $\begingroup$ That doesn't explain why the OP's work is incorrect though. $\endgroup$ – user370967 May 25 '17 at 13:35

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