Let $f(x)=|x|+|x-1|$. Check the differentiability of $f$ at $x=0$ and $x=1$ 
Let $f(x)=|x|+|x-1|$. Check the differentiability of $f$ at $x=0$ and
  $x=1$

The short answer would be since $|x|$ is not differentiable at $x=0$ and $|x-1|$ is not differentiable at $x=1$, so $f(x)$ will not be differentiable at $x=0,1$ 
{Is it valid to claim this in every case ?}
Long answer:
$f(x)= \begin{cases} 1 - 2x & \text{ if } x< 0 \\ 1 & \text{ if } 0 \leq x < 1 \\ 2x -1 & \text{ if }x\geq1 \end{cases}$ 
$\lim_{x\to 0^{+}} \frac{f(x) - f(0)}{x-0}  = \lim_{x\to 0^+} \frac{1-1}{x-0} = 0$
$\lim_{x \to 0^-}\frac{f(x) - f(0)}{x-0}=\lim_{x \to 0^-} \frac{1-2x}{x-0}=  -\infty$ is not finite. So not differentiable at 0
$\lim_{x\to 1^+}\frac{f(x) - f(1)}{x-1}= \lim_{x\to 1^+}\frac{2(x-1)}{x-1} =2$
$\lim_{x \to  1^-}\frac{f(x) - f(1)}{x-1}=\lim_{x\to 1^-}\frac{1-1}{x-1}=0$
which are unequal and hence not differentiable at 1.
Is this proof fine ?
 A: No, it is not fine:$$\lim_{x\to1^-}\frac{f(x)-f(0)}{x-0}=\lim_{x\to1^-}\frac{1-2x-1}x=-2\neq-\infty.$$The rest is fine.
However, consider the functions $f(x)=|x|$ and $g(x)=-|x|$. Neither of them is differentiable at $0$. However, their sum is. Do you see why I am writing this?
A: Except for the mistake that I pointed out in a comment above, yes your long answer seems good. You've shown that $\lim_{x\to a^+}\frac{f(x) - f(a)}{x-a}$ and $\lim_{x\to a^-}\frac{f(x)-f(a)}{x-a}$ are different for the relevant $a$, and therefore $\lim_{x\to a}$ doesn't exist in either of those cases, and the function is by definition not differentiable.
Your short answer is flawed, however. Specifically, if you have a differentiable function $f$, and a non-differentiable function $g$, then the sum of $g$ and $(f-g)$ is differentiable. This construction works in general as a counterexample for several similar statements of the kind "the sum of two nice things is a nice thing implies that the sum of two non-nice things is a non-nice thing", like:


*

*The sum of two non-differentiable functions is non-differentiable

*The sum of two discontinuous functions is discontinuous

*The sum of two irrational numbers is irrational

*The sum of two transcendental numbers is transcendental

A: The short answer is almost right: since $|x|$ is not differentiable at $x=0$ and $|x-1|$ is differentiable at $x=0$, it follows that $f(x)$ can't be differentiable. This is because if it was, $|x|=f(x)-|x-1|$ would be the difference of two differentiable functions, so differentiable.
Your long version is good apart from a mistake in evaluating one of the limits (which should be $-2$ not $-\infty$).
A: You can simply get them by plotting the graph of the function. The function is non-differentiable at all sharp points and hence at $x=0,1$

