# Squeeze theorem 2 bounded functions

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

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

• You should have gotten rid of this factor you have at the end instead of getting rid of the $\sin y$ – Ninad Munshi Mar 3 at 9:51
• There exist no limits. – haidangel Mar 3 at 9:55

## 2 Answers

$$-1 \leq \sin y \leq 1$$ 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.

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

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