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  $\sin(x^2 + y^2) \approx x^2 + y^2$?
 A: $$ \lim_{(x,y) \to (0,0)} \dfrac{\sin(x^2 + y^2)}{x^2 + y^2}  = 1
\iff \left[\forall r>0\ \ \ \exists \delta > 0\ \ \ 
d((x,y),(0,0))<\delta\implies
\left| \dfrac{\sin(x^2 + y^2)}{x^2 + y^2} - 1\right|\le r\right]\\
\iff \left[\forall r>0\ \ \exists \delta > 0\ \ \ 
x^2+y^2<\delta^2\implies
\left| \dfrac{\sin(x^2 + y^2)}{x^2 + y^2} - 1\right|\le r\right]\\
\iff \left[ \forall r>0\ \ \exists \delta > 0\ \ \ 
0\le U<\delta^2\implies
\left| \dfrac{\sin(U)}{U} - 1\right|\le r\right]\\
\iff \lim_{U \to 0^+} \dfrac{\sin(U)}{U}  = 1
$$which is true because $\sin'(0) = \cos(0) = 1$.
But both $x$ and $y$ must be close to zero.
A: Since $(x, y) \rightarrow (0,0)$ consider $r^2=x^2+y^2 $ then we can consider the limit of $\frac {\sin (r^2)}{r^2}$, so we will consider what happens when $r\to 0$
$\lim _ {r\to 0}\frac {\sin (r^ 2 )}{r^2}=\frac {0}{0}$,
Now we apply l'hospitals rule, differentiating the top and bottom to get:
$\lim _ {r\to 0}\frac {\sin (r^ 2 )}{r^2}=\lim _ {r\to 0}\frac {2r\cos(r^ 2 )}{2r}=\lim _ {r\to 0}\cos(r^ 2 )=1 $.
A: It's not even true that $\sin(x^2 + y^2) \approx x^2 + y^2$ if $x$ is close to $0$, because $y$ could be quite large.  What is true is that if $x^2+y^2$ is small, then $\sin(x^2 + y^2) \approx x^2 + y^2$.  But in any case, even if you said that it wouldn't be a very rigorous limit proof, because you're just talking about approximations.  But it could be the start of the proof if you take the Taylor series of $sin(x^2+y^2)$ and take the limit.  Or, more simply, you can take @Dane's suggestion and do $r^2 = x^2+y^2$.
