Does $\lim_{(x,y)\to(0,0)}[x\sin (1/y)+y\sin (1/x)]$ exist?

Question

This is an exercise from my calculus class. The function is defined as $x\sin (1/y)+y\sin (1/x)$ if $x\neq0 $ and $y\neq0 $, and $0$ if $x=0 $ or $y=0$.

I'm pretty confident the limit exists and should be $0$, because: $$\lim_{(x,y)\to(0,0)}[x\sin (1/y)+y\sin (1/x)]=\lim_{(x,y)\to(0,0)}[x\sin (1/y)]+\lim_{(x,y)\to(0,0)}[y\sin (1/x)]$$

And: $x\leq x\sin(1/y)\leq x$, so $\lim_{(x,y)\to(0,0)}[x\sin (1/y)]=0$ right? (The same can be said for $\lim_{(x,y)\to(0,0)}[y\sin (1/x)]$)

However, I tried checking my answer, and according to Wolfram Alpha the limit doesn't exist. Is this because I'm wrong, or is it just because $x\sin (1/y)+y\sin (1/x)$ is undefined for $y\neq0 $ and $y\neq0 $

Answer 1

For $x\ne0$ and $y\ne0$ we have $$ \left|x\sin\frac{1}{y}+y\sin\frac{1}{x}\right|\le \left|x\sin\frac{1}{y}\right|+\left|y\sin\frac{1}{x}\right|\le|x|+|y| $$ So, for $$ f(x,y)=\begin{cases} x\sin\dfrac{1}{y}+y\sin\dfrac{1}{x} & \text{if $x\ne0$ and $y\ne0$} \\ 0 & \text{if $x=0$ or $y=0$} \end{cases} $$ we have $$ |f(x,y)|\le |x|+|y| $$ for all $(x,y)$. Therefore $$ \lim_{(x,y)\to(0,0)}f(x,y)=0 $$ by the squeeze theorem.

Be careful that $x\sin(1/y)\le x$ is not true in general, but you just need the absolute value and $|x\sin(1/y)|\le|x|$ is true (provided $y\ne0$, of course).

WolframAlpha is a great resource, but it doesn't always tell the truth. ;-)

Answer 2

For $x,y\neq 0,$ $$|f(x,y)-f(0,0)|\leq |x|+|y|\leq 2\sqrt{x^2+y^2}.$$ Now choosing $\delta=\epsilon/2,$ we get $$|f(x,y)-f(0,0)|<\epsilon,$$ whenever $\sqrt{x^2+y^2}<\delta.$ Therefore $\displaystyle\lim_{(x,y)\rightarrow(0,0)}x\sin{1/y}+y\sin{1/x}=0.$