Partial derivative $f(x,y)=x^2 \sin y$

Question

I was studying when I found this partial derivative: $$z=f(x,y)=x^2 \sin y$$

$$ fy= \lim\limits_{\Delta \to 0}\frac{\Delta yz}{\Delta y} = \lim\limits_{\Delta \to 0}\frac{x^2 \sin(y+\Delta y) - x^2 \sin y}{\Delta y} = \lim\limits_{\Delta \to 0}\frac{x^2\sin y + x^2 \sin(\Delta y)\cos y - x^2 \sin y}{\Delta y} = \dots = x^2\cos y $$

Knowing $ \sin(A+B)=\sin A \cos B + \cos A\sin B$, I can't see where did $ \cos \Delta y$ go.

Answer 1

Since $\cos(\cdot)$ is a continuous function in a neighborhood of $0$,

$$\lim_{\Delta\to 0} x^2 \cos (\Delta) \sin (y)= x^2 \cos(0) \sin (y)= x^2 \sin (y)$$