<s>[INST] Show that $\lim_{(x,y) \to (0,0)} {\frac{x\sin y - y\sin x}{x^2 +y^2}}$ does not exist Show that $\lim_{(x,y) \to (0,0)} {\frac{x\sin y - y\sin x}{x^2 +y^2}}$ does not exist
I did use Wolfram Alpha and it says this limit does not exist.
I'm trying to prove this with sequential definition of multivariable function. So basically, I have to find two sequences $(u_k)$ and $(v_k)$ such that they approach $(0,0)$ but the two sequences $f(u_k)$ and $f(v_k)$ approach two different limits. I tried various things but nothing works out. Could you give me some hint about this problem? Thank you in advance!
 [/INST] Taylor expansion to the rescue:
$$
\begin{split}
\frac{x\sin y - y \sin x}{x^2 + y^2}&=\frac{1}{x^2+y^2}\left( 
x\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}y^{2n+1} - 
y\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}
\right) \\
&=\frac{1}{x^2+y^2}\left( 
xy\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}y^{2n} - 
xy\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n}
\right)\\
&=\frac{xy}{x^2+y^2}\left( 
\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}y^{2n} - 
\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n}
\right)\\
&=\frac{xy}{x^2+y^2}\left( 
1-\frac{y^2}6+\sum_{n=2}^\infty \frac{(-1)^n}{(2n+1)!}y^{2n} -
1+\frac{x^2}6-\sum_{n=2}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n}
\right)\\
&=\frac{xy}{x^2+y^2}\left(
\frac{x^2-y^2}6
+\sum_{n=0}^\infty \frac{(-1)^n}{(2n+5)!}y^{2(n+2)} -
\sum_{n=0}^\infty \frac{(-1)^n}{(2n+5)!}x^{2(n+2)}
\right)\\
&=\frac{xy}{x^2+y^2}\left(
\frac{x^2-y^2}6
+y^4\sum_{n=0}^\infty \frac{(-1)^n}{(2n+5)!}y^{2n} -
x^4\sum_{n=0}^\infty \frac{(-1)^n}{(2n+5)!}x^{2n}
\right)
\end{split}
$$
Therefore:
$$
\begin{split}
\lim_{(x,y)\to(0,0)}\frac{x\sin y - y \sin x}{x^2 + y^2}&=
\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}\left(
\frac{x^2-y^2}6 +\mathcal O(x^4) +\mathcal O(y^4) \right)\\&=
\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}\frac{x^2-y^2}6 = 0.
\end{split}
$$
Note that while the $\frac{xy}{x^2+y^2}$ part takes values from $-\frac12$ to $\frac12$ around $(0,0)$, the $x^2-y^2$ part goes to zero no matter how does one approach $(0,0)$. Therefore, the whole thing also goes to zero.
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