How to prove that this limit doesn't exits? The limit is $$ \lim_{(x,y)\to(0,0)} \frac{x\sin(y)-y\sin(x)}{x^2 + y^2}$$
My calculations: I substitute $y=mx$
\begin{align}\lim_{x\to 0} \frac{x\sin(mx)-mx\sin(x)}{x^2 + (mx)^2} &= \lim_{x\to 0} \frac{x(\sin(mx)-m\sin(x)}{x^2(1 + m^2)}\\ &= \lim_{x\to 0} \frac{1}{1+m^2}\bigg[\frac{\sin(mx)}{x}- \frac{m\sin(x)}{x}\bigg]\end{align}
Can I say that the limit $$ \lim_{x\to 0}\frac{\sin(mx)}{x}$$
doesn't exist because it depends on $m$, so the entire limit doesn't exist?
 A: No, you cannot. The limit$$\lim_{x\to0}\frac{\sin(mx)}x$$does exist. It is equal to $m$.
But, and that's more importante, the limit$$\lim_{x\to 0} \frac{1}{1+m^2}\bigg[\frac{\sin(mx)}{x}- \frac{m\sin(x)}{x}\bigg]$$is equal to $0$ for every $m$.
A: I would just
throw in the
first terms of
the power series
and see what happens.
Since
$\sin(x)
= x-x^3/6+O(x^5)
$,
\begin{align}
\frac{x\sin(y)-y\sin(x)}{x^2 + y^2}
&=\frac{x(y-y^3/6+O(y^5))-y(x-x^3/6+O(x^5))}{x^2 + y^2}\\
&=\frac{xy-xy^3/6+O(xy^5)-yx+yx^3/6+O(x^5y))}{x^2 + y^2}\\
&=\frac{-(xy^3+yx^3)/6+O(xy^5)+O(x^5y))}{x^2 + y^2}\\
&=-\frac{xy(y^2+x^2)/6+O(xy^5)+O(x^5y))}{x^2 + y^2}\\
&=-\frac{xy}{6}+\frac{xy(O(y^4)+O(x^4))}{x^2 + y^2}
\end{align}
so
\begin{align}
\left|\frac{x\sin(y)-y\sin(x)}{x^2 + y^2}+\frac{xy}{6}\right|
&=\left|\frac{xy(O(y^4)+O(x^4))}{x^2 + y^2}\right|\\
&\le O(y^4)+O(x^4)\\
\end{align}
since
$0 \le (|x|-|y|)^2
=x^2+y^2-2|xy|$
so
$\frac{|xy|}{x^2+y^2}
\le \frac12$.
Therefore
the limit is zero.
A: No, we have $$\lim_{x\rightarrow 0} \frac{\sin(mx)}{x}=\lim_{x\rightarrow 0}m\cdot \frac{\sin(mx)}{mx}=m=\lim_{x\rightarrow 0}\frac{m\sin(x)}{x}$$ , so the limit you want to calculate is $0$ , no matter what $m$ is.
