<s>[INST] A proof for $\lim_{(x,y) \to (0,0)} \frac{\sin(x^2 + y^2)}{x^2 + y^2}$ To prove the following limit $ \lim_{(x,y) \to (0,0)} \dfrac{\sin(x^2 + y^2)}{x^2 + y^2}  = 1$ is it sufficient to say that around $(x,y) \to (0,0) $ we have  $\sin(x^2 + y^2) \approx x^2 + y^2$?
 [/INST] $$ \lim_{(x,y) \to (0,0)} \dfrac{\sin(x^2 + y^2)}{x^2 + y^2}  = 1
\iff \left[\forall r>0\ \ \ \exists \delta > 0\ \ \ 
d((x,y),(0,0))<\delta\implies
\left| \dfrac{\sin(x^2 + y^2)}{x^2 + y^2} - 1\right|\le r\right]\\
\iff \left[\forall r>0\ \ \exists \delta > 0\ \ \ 
x^2+y^2<\delta^2\implies
\left| \dfrac{\sin(x^2 + y^2)}{x^2 + y^2} - 1\right|\le r\right]\\
\iff \left[ \forall r>0\ \ \exists \delta > 0\ \ \ 
0\le U<\delta^2\implies
\left| \dfrac{\sin(U)}{U} - 1\right|\le r\right]\\
\iff \lim_{U \to 0^+} \dfrac{\sin(U)}{U}  = 1
$$which is true because $\sin'(0) = \cos(0) = 1$.
But both $x$ and $y$ must be close to zero.
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