Q: Continuity and Differentiability at end point of an interval Let $ f(x) = x(\sqrt {x} + \sqrt{x+1})  $
The problem had asked to check continuity and Differentiability at $ x = 0 $
First of all I noticed that the Natural Domain of the function is non negative Real Numbers i.e zero is the least point of interval
Secondly I noted that $f(0)=0 $
And that the Right Hand Limit is also zero , implying that $f(x)$ is continuous at $0$.
My lecturer had told that to check Differentiability on the end point of an interval we need that the Right Hand Derivative must be finite . When I evaluate the  Right Hand Derivative it comes to be equal to $1$ (and hence finite, and therefore differentiable )
$$  \lim_{h\to 0} \frac {f(0+h) - f(0)}{h} (where  h>0) $$
Which evaluates to $$ \lim_{h\to 0} \frac {h(\sqrt {h} + \sqrt {1+h}) - 0}{h} , $$ the h cancels out and we are left with 1 .
But the problem book says that the function is not differentiable at $x=0$ as it not defined from the Left side . I am confused on which is correct ?

A: All your calculations are correct, so this comes down to definitions. Usually, the definition of differentiability requires the function to be defined at some neighborhood of a point. This is not true for $x=0$. For example, the Wikipedia definition talks about differential functions as something that exists for points in an open set on which $f$ is defined. There is no open set on which your function is defined that would include $0$, so the function is not differentiable at $0$.
This means that you have a slightly strange situation here where the function has a derivative at $0$, but is not differentiable at $0$.
