4
$\begingroup$

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 $

$\endgroup$
  • 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
8
$\begingroup$

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. ;-)

$\endgroup$
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.