$\mathbf{Question:}$
Prove that
$\displaystyle \lim_{(x,y)\to (0,0)} \dfrac{\sin(2x+2y)-2x-2y}{\sqrt{x^{2}+y^{2}}}=0$
$\mathbf{My\ ideas:}$
I will use the First Order Approximation Theorem.
But how can I apply this?
Please show me understanbly clear because this is my first example related to the theorem. I want to learn. Thank you:)
Hint: the gradient $\nabla f(0,0)$ of $f(x,y)=\sin(2x+2y)$ at $(0,0)$ is the vector
$$\nabla f(0,0)=(2,2), $$
as $\frac{\partial f}{\partial x}=2\cos(2x+2y)=\frac{\partial f}{\partial y}=2\cos(2x-2y).$
$f$ is differentiable at $(0,0)$ (which implies $f$ continuous at $(0,0)$) if
$$\lim_{(x,y)\rightarrow (0,0)} \frac{f(x,y)-f(0,0)-\langle \nabla f(0,0),(x,y)\rangle}{\sqrt{x^2+y^2}}, $$
i.e.
$$\lim_{(x,y)\rightarrow (0,0)} \frac{f(x,y)-f(0,0)-\langle(2,2),(x,y)\rangle}{\sqrt{x^2+y^2}}, $$
or
$$\lim_{(x,y)\rightarrow (0,0)} \frac{\sin(2x+2y)-2x-2y}{\sqrt{x^2+y^2}}, $$
which is the limit you consider. I used the fact that
$$f(0,0)=0,$$ $$\langle \nabla f(0,0),(x,y)\rangle=2x+2y,$$ $$\|(x,y)\|=\sqrt{x^2+y^2}. $$
To compute such limit, consider it along the $x$-axis, for all points $(x,0)$ going to $(0,0)$, or the $y$-axis (all points $(0,x)$ going to $(0,0)$). Can you compute it in these 2 special cases?