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$$\lim_{(x,y) \to (0,0)} \frac{x^2 \sin (y)}{x^2+y^2} $$

Squeeze Theorem:

$$g(x) \leq f(x) \leq h(x)$$

$$-1 \leq \sin(y) \leq 1$$

$$- \frac{x^2}{x^2+y^2}\leq \frac{x^2 \sin (y)}{x^2+y^2} \leq \frac{x^2}{x^2+y^2} $$ Also $$ 0 \leq \frac{x^2}{x^2+y^2} \leq 1 $$ What should be my next step How does this comes out to be zero

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  • $\begingroup$ You should have gotten rid of this factor you have at the end instead of getting rid of the $\sin y$ $\endgroup$ – Ninad Munshi Mar 3 at 9:51
  • $\begingroup$ There exist no limits. $\endgroup$ – haidangel Mar 3 at 9:55
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$-1 \leq \sin y \leq 1$ is not helpful in this case.

Use the fact that $|\sin y | \leq |y|$ so $-|y| \leq \sin y \leq |y|$. Now you get the bounds $\pm \frac {x^{2}|y|} {x^{2}+y^{2}}$. If you note that $\frac {x^{2}} {x^{2}+y^{2}} \leq 1$ you get $-|y| \leq f(x,y) \leq |y|$ and you can now apply Squeeze Theorem.

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I would use polar coordinates and equivalence of functions: $$\frac{x^2 \sin (y)}{x^2+y^2}=\frac{r^2\cos^2\theta\sin(r\sin\theta)}{r^2}\sim_{r\to 0}r\sin\theta\cos^2\theta.$$

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  • $\begingroup$ Then, no limits $\endgroup$ – haidangel Mar 3 at 12:10
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    $\begingroup$ @haidangel: not at all – $\:r$ tends to $0$ and $\sin\theta\cos^2\theta$ is bounded. $\endgroup$ – Bernard Mar 3 at 12:20
  • $\begingroup$ Its easier to do using polar coordinates but i had to do it using Squeeze theorem which is where i was getting stuck $\endgroup$ – Muskan Mar 3 at 12:27
  • $\begingroup$ @Muskan: In this case, yes. What makes things still easier (or at least shorter) is also the use of some elementary asymptotic analysis. $\endgroup$ – Bernard Mar 3 at 12:30

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