<s>[INST] How to prove the given function is not differentiable analytically? 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 $x=2$?
 [/INST] $$f (2^-)=\lim_{x\to 2^-} (\frac {x}{2}+2)=1+2=3$$
$$ f (2^+)=\lim_{2^+}\sqrt {2x}=2$$
$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 $$\lim_{x\to 2}\epsilon (x)=0$$
</s>