limits of the differential of $\sqrt{1-\cos x}$

Question

How do I find the left and right limits as $x$ goes to $0$ of the differential of the function $y = \sqrt{1 - \cos(x)}$? When I put the equation into Wolfram Alpha, the left limit was $- 1/\sqrt2$ and the right limit was $1/\sqrt{2}$ but I don't understand how these are the answers.

Answer 1

$$y=\sqrt{1-\cos x}$$ $$\frac{dy}{dx}=\frac{\sin x}{\sqrt{1-\cos x}}$$ clearly from this we can see that $\cos x<1$ so $x>0$