Calculating $\lim_{(\Delta x,\Delta y)\to(0,0)} \frac{\Delta x(\sin (\Delta y) -\Delta y)}{\sqrt{((\Delta x)^2+(\Delta y)^2})^3} $

Question

I'm trying to calculate $$\displaystyle\lim_{(\Delta x,\Delta y)\to(0,0)} \frac{\Delta x(\sin (\Delta y) -\Delta y)}{\sqrt{((\Delta x)^2+(\Delta y)^2})^3} $$ to tell if the following function is differentiable at $(0,0)$: $f(x,y)=\frac{x(\sin y-y)}{x^2+y^2} $ is $(x,y)\neq (0,0)$, else $0$. I was trying to determine this using the definition of differentiability $f(x_0+\Delta x, y_0+\Delta y)-f(x_0,y_0)=f_x(x_0,y_0)\Delta x +f_y(x_0,y_0)\Delta y +\epsilon\sqrt{(\Delta x)^2+(\Delta y)^2} $ where $\epsilon(\Delta x,\Delta y)\to 0$ when $\Delta x,\Delta y\to 0$.

I would be happy to hear your thoughts. Thanks in advance :)

Answer 1

Note that

$$\frac{x(\sin (x-y))}{\sqrt{(x^2+y^2)^3}}=\frac{\sin (x-y)}{x-y}\frac{x(x-y)}{\sqrt{(x^2+y^2)^3}}\to 0$$

indeed

$$\frac{\sin (x-y)}{x-y}\to 1$$

and

$$\frac{x(x-y)}{\sqrt{(x^2+y^2)^3}}=\frac{\rho^2(\cos^2 \theta-\sin \theta\cos \theta)}{\rho\sqrt{\rho}}=\sqrt{\rho}(\cos^2 \theta-\sin \theta\cos \theta)\to 0$$