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

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 $y\neq0$

• Inequality $x\leq x\sin(1/y)\leq x$ is not correct. Instead you could write $|x\sin(1/y)|\leq |x|$ and conclude. Wolfram doesn't give you a limit because this function is not defined at $(0,0)$ (edit if you like the last part of your question) Commented Mar 28, 2016 at 21:54

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. ;-)
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 $$\displaystyle\lim_{(x,y)\rightarrow(0,0)}x\sin{1/y}+y\sin{1/x}=0.$$