Differentiability of piecewise function using the definition analytically In the following example particularly 
$$
f(x)= \begin{cases}  (x-2)^2  + 5 \quad\text{when $x\geq 2$} \\ 
   (x-2)^2  + 4 \quad\text{when $x&lt;2$} \end{cases}
$$
for the above function I know quite well graphically that the function is not differentiable as the when lim $h\to 0^-$ of difference quotient is approaching infinity.

But my question now is how to prove that analytically without graph (animation). In other words if we differentiate LHS and RHS at $x=2$ we will find the same value which is $2(x-2)=0$

For example for $f(x)=|x|$ in many videos on the internet, the tutor differentiate both sides independently but if we do this in my example it will yield the same real number which is zero though the left hand limit shouldn't exist?
$$ \frac{d}{dx}f(2)=\lim_{h\to 0} \frac{f(2+h)-f(2)}{h} $$
From the left of $0$, we have
$$ \lim_{h\to 0^-} \frac{(2+h-2)^2+4-(2-2)^2-5}{h} $$ 
$$ =\lim_{h\to 0^-} \frac{h^2-1}{h}= \lim_{h\to 0^-} h-\frac{1}{h} =\infty $$ 
And now from the right of $0$, we have
$$ \lim_{h\to 0^+} \frac{(2+h-2)^2+5-(2-2)^2-5}{h} $$ 
$$ =\lim_{h\to 0^+} \frac{h^2}{h}= \lim_{h\to 0^+} h =0 $$ 
Therefore, $f(x)$ is not differentiable at $x=2$.
Whether or not this will help you depends on what the exact posing of the problem is, but you might find it helpful to recall that a function that is not continuous at a point cannot be differentiable at that point.