# function with two variables

find the value of the nexg limit:

$$\lim_{(x,y) \to (0,0) }\frac{sin(x^3+y^3)}{x^2+y^2}$$

I have tried to use $$1\ge |sinx|$$ but it didn't work.

## 1 Answer

Transform to polar coordinates by setting: $$x=r\cos\theta$$ $$y=r\sin\theta$$

Using $|\sin (u)|≤|u|$ we get:

$$\left | \frac{\sin(x^3+y^3)}{x^2+y^2} \right |=\left |{\sin(r^3(\cos^3 \theta + \sin ^3 \theta)) \over r^2}\right |≤|r(\cos ^3 \theta + \sin ^3 \theta)|≤2r$$

Which tends to $0$ as $r \to 0^+$. Hence the desired limit equals $0$.

• thank you for your answer but why $$r \to 0^+$$? – BlessYour Karma Jun 20 '15 at 12:23
• @BlessYourKarma The vector $(x,y)$ tends to $0$ if and only if the radius gets arbitrarily small, that is $r \to 0^+$ (it tends to zero from the right because the radius is always positive by definition). To see this, you can note that: $|(x,y)|=\sqrt{x^2+y^2}=\sqrt{r^2(cos^2 \theta + \sin ^2 \theta)}=r$. Is this what you're referring to? – Reveillark Jun 20 '15 at 12:53