Is the piecewise-defined function differentiable The function is defined as 
$$f(x)=\begin{cases}x^2, &\text{ for }x\leq 1\\ \sqrt{x}, &\text{ for }x>1\end{cases}$$ and
Is this function differentiable at $x=1$?
I thought that since $\lim_{x\to 1}$ of $f'(x)$ exists then it IS differentiable. And I think this limit does exist so it should be differentiable. Book says no. My logic must not be correct here.

Taking $\lim\limits_{x\to 1}f'(x)$ is not well defined, until we are convinced that $f'(x)$ exists. If $f'(1)$ does exist, we will have that the left and right hand limits defining the derivative are equal. If we denote $f'_-(x)$ and $f'_+(x)$ as the left and right derivatives respectively, we get that \begin{align*} f'_-(x) &= 2\\
f'_+(x) &=\frac{1}{2} \end{align*}
The left and right derivatives exist, but they are not equal. Thus, $f$ is not differentiable.
