Show that $f(x,y)=\sin{(x+y)}$ is differentiable at $(1,1)$

Question

I'm trying to show this using the definition for differentiability in $\mathbb{R}^2:$

A function $f(x,y)$ of two variables is differentiable in $(x,y)=(a,b)$ iff $ \ \exists \ $ constants $A,B:$

$$f(1+h,1+k)-f(a,b)=Ah+Bk+\sqrt{h^2+k^2} \ \rho(h,k),$$ and $\rho(h,k)\rightarrow 0$ as $(h,k)\rightarrow(0,0)$.

I get that

$$f(1+h,1+k)-f(1,1)=\sin(2+h+k)-\sin{2}=0h+(-\sin{2})k+\sin{(2+h+k)} = \\ =Ah+Bk+\sqrt{h^2+k^2}\rho(h,k).$$

Identifying sides I get that $A=0, \ B=-\sin{2}$ and $\rho(h,k)=\sin{(2+h+k)}/\sqrt{h^2+k^2.}$

So the constants $A$ and $B$ are ok, but the trouble now is showing that $$\lim_{(h,k)\rightarrow(0,0)} \frac{\sin{(2+h+k)}}{\sqrt{h^2+k^2}} =0.$$

Answer 1

$$Df(x_0,y_0)= f_x(x_0,y_0)dx + f_y(x_0,y_0)dy \implies$$

$$ Df(1,1)= cos(2)dx+ cos(2)dy = cos(2)(dx+dy) $$

$$ f(1+dx,1+dy) -f(1,1)-Df(1,1)=$$ $$sin(2+dx+dy) - sin(2) - cos(2)(dx+dy)=$$

$$sin(2)cos(dx+dy)+cos(2)sin(dx+dy) -sin(2)-cos(2)(dx+dy) = $$ $$sin(2)[cos(dx+dy)-1]sin(2)+cos(2)[sin(dx+dy)-(dx+dy)]$$

Note that $$ | sin(2)[cos(dx+dy)-1]sin(2)+cos(2)[sin(dx+dy)-(dx+dy)]/ \sqrt {(dx)^2 + (dy)^2}\to 0 $$

That is $f(x,y)= sin(x+y)$ is differentiable at $(1,1)$.