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!
 A: For $x,y\ne 0$, we have:
$$\frac{x\sin y-y\sin x}{x^2+y^2}=\left(\frac{\sin y}{y}-\frac{\sin x}{x}\right)\frac{xy}{x^2+y^2}$$
so when $(x,y)\to (0,0)$, the first factor $\frac{\sin y}{y}-\frac{\sin x}{x}$ converges to $1-1=0$ and the second factor is bounded:
$$|xy|\le\frac{1}{2}(x^2+y^2)\implies\left|\frac{xy}{x^2+y^2}\right|\le \frac{1}{2}$$
so the whole function converges to zero. The same happens on the lines $x=0$ or $y=0$ (as the function is zero there).
So, it seems to me that $\lim_{(x,y)\to(0,0)}\frac{x\sin y-y\sin x}{x^2+y^2}$ exists after all, and it is $0$.
A: 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.
A: Taylor's remainder theorem states that a function differentiable at $0$ can be written as
$$
f(x)=f(0)+f'(0)x+h_1(x)x
$$
where $\lim_{x\rightarrow 0}h_1(x)=0$.  Sometimes, this is written as
$$
f(x)=f(0)+f'(0)+\frac{f''(c)}{2}x^2
$$
for some $c$ in $(0,x)$, but the first formulation will be enough for our purposes.  Specializing to $\sin$, we have
$$
\sin(x)=x+h_1(x)x,
$$
where $\lim_{x\rightarrow 0}h_1(x)=0$.
For the problem of interest, we use the polar transformation $x=r\cos(\theta)$ and $y=r\sin(\theta)$ to get
$$
\lim_{(x,y)\rightarrow (0,0)}\frac{x\sin(y)-y\sin(x)}{x^2+y^2}=
\lim_{r\rightarrow 0^+}\frac{r\cos(\theta)\sin(r\sin(\theta))-r\sin(\theta)\sin(r\cos(\theta))}{(r\cos(\theta))^2+(r\sin(\theta))^2}.
$$
The denominator simplifies to $r^2$ through the trigonometric identity $(\sin(\theta))^2+(\cos(\theta))^2=1$.  For the numerator, we use the Taylor expansion to get that this limit is
$$
\lim_{r\rightarrow 0^+}\frac{r^2\cos(\theta)\sin(\theta)(1+h_1(r\sin(\theta)))-r^2\cos(\theta)\sin(\theta)(1+h_1(r\cos(\theta)))}{r^2}.
$$
The $r^2$'s cancel as well as terms in the numerator, so this simplifies to
$$
\lim_{r\rightarrow 0^+}\cos(\theta)\sin(\theta)(h_1(r\cos(\theta))-h_1(r\sin(\theta))).
$$
Since $\cos(\theta)$ and $\sin(\theta)$ are bounded between $-1$ and $1$ their product is also between $-1$ and $1$, so we can squeeze this limit between the following two limits:
$$
-\lim_{r\rightarrow 0^+}(h_1(r\cos(\theta))-h_1(r\sin(\theta)))\leq
\lim_{r\rightarrow 0^+}\cos(\theta)\sin(\theta)(h_1(r\cos(\theta))-h_1(r\sin(\theta)))\leq \lim_{r\rightarrow 0^+}(h_1(r\cos(\theta))-h_1(r\sin(\theta))).
$$
We know that $-r\leq r\cos(\theta)\leq r$ and $-r\leq r\sin(\theta)\leq r$ since $\cos(\theta)$ and $\sin(\theta)$ are both bounded between $-1$ and $1$.  Therefore, using the squeeze theorem again,
$$
-\lim_{r\rightarrow 0^+} r\leq\lim_{r\rightarrow 0^+}r\cos(\theta)\leq\lim_{r\rightarrow 0^+}r
$$
and
$$
-\lim_{r\rightarrow 0^+} r\leq\lim_{r\rightarrow 0^+}r\sin(\theta)\leq\lim_{r\rightarrow 0^+}r.
$$
Since $\lim_{r\rightarrow 0^+}r=0$, all of the limits above exist and are equal to $0$.
Putting this all together, since $r\cos(\theta)$ and $r\sin(\theta)$ approach $0$ as $r$ approaches $0$ and $\lim_{x\rightarrow 0}h_1(x)=0$, we conclude that the arguments to the two copies of $h_1$ in the following expression are approaching $0$.  Using the limit property that we know about $h_1$, we conclude that $h_1(r\cos(\theta))$ and $h_1(r\sin(\theta))$ both approach $0$, that is
$$
\lim_{r\rightarrow 0^+}(h_1(r\cos(\theta))-h_1(r\sin(\theta)))=0.
$$
By working backwards, we can see that the original limit is $0$.
