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 $

  • 1
    $\begingroup$ 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) $\endgroup$ – Nikolaos Skout Mar 28 '16 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. ;-)

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    $\begingroup$ Liked the warning about "WolframAlpha". People should not use it to solve exercises like these. Rather it should be used for computationally intensive problems. +1 $\endgroup$ – Paramanand Singh Mar 30 '16 at 7:51

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