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.
On
Well, $\lim_{x \to 1^-} f'(x) = 2$ and $\lim_{x \to 1^+} f'(x) = \frac{1}{2}$, so $\lim_{x \to 1} f'(x) $ doesn't exist.
But to really prove that $f$ is not differentiable in $1$, try to use the definition.
On
Generally, if you graph a piecewise function and at any point it doesn't look "smooth" (there's a "sharp" turn), then it is not differentiable at that point.
More rigorously, the derivatives of the two parts of the function are not the same at $1$, so it is not differentiable. Specifically, $$\frac{d}{dx}x^2 = 2x = 2 \text{ when $x=1$}$$ and $$\frac{d}{dx}\sqrt x = \frac1{2\sqrt x} = \frac12 \text{ when $x=1$}$$ and clearly, $2\neq\frac12$. Thus the derivative at 1 doesn't exist.
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.