I was watching some video about differentiability at some point and I was presented with this function: $f(x)=\left\{ \begin{aligned} \frac{1}{2}x,\quad x < 1\\ \sqrt{x}-1 ,\quad x\ge1\\ \end{aligned} \right.$
And the left hand derivative and right hand derivative, they are both equal to $\frac{1}{2}$ at $x=1$. So that would mean that the derivative exist, but we also know that the function is discontinuous which implies that the function is not differentiable at that point.
So my question is, if we know that differentiabiliy implies continuity,why can't we apply that theorem here. If we tried to find existence of the derivative first at the point $x=1$, which later shows us that left and right side are both equal to $\frac{1}{2}$ then that would imply continuity, but we see the function is discontinuous.
Where is the problem in my logic?