# Difficulty proving multivariate limit involving sin^2(x) does not exist

I am attempting to prove that $$\lim_{(x,y) \to (0,0)} \frac{x \sin^{2}(y)}{x^{2} + y^{2}}$$ does not exist, but am having issues. I can't seem to find a path for which the value is not zero (to prove that the limit value depends on the given path) , however I do believe that it does not exist, as I can't seem to prove that it does either with the Squeeze theorem.

I have attempted approaching along $y=0$, $x=0$, $y=mx$, $y=mx^{2}$ to name a few, and the usual strategies of attempting to cancel out factors or splitting up my equation don't seem to help as they all just give me zero. Am I on the wrong track here with believing the limit does not exist in the first place?

## 3 Answers

Hint: What if it were $y^2$ instead of $\sin^2 y$ in the numerator?

• Ahh, right the limit would approach zero. Thanks. – HavelTheGreat Jun 4 '15 at 22:58
• Small suggestion: The limit doesn't approach anything, either it exists or it doesn't, and if it does, it's just a number and doesn't dance around or approach anything. – zhw. Jun 4 '15 at 23:05
• Right, complete misuse of 'approach' on my part. Thanks again. – HavelTheGreat Jun 4 '15 at 23:11

$x^2+y^2 \geq 2|x||y| \to \left|\dfrac{x\sin^2y}{x^2+y^2}\right| \leq \dfrac{|x|\sin^2y}{2|x||y|} \leq \dfrac{1}{2}|\sin y|\cdot \left|\dfrac{\sin y}{y}\right| \to$ limit $= 0$ by squeeze theorem.

How about using polar coordinates? Define $$x^{2}+y^{2}=r$$ , $$x=r\cos(\beta)$$, $$y=r\sin(\beta)$$, Substitute to get $$\lim_{r\to 0}\frac{r\cos(\beta)\sin^{2}((r\sin(\beta))}{r}$$ The outer $$r$$ cancels out with the denominator and the other $$r$$ cause the function approaching to $$0$$.