# Thread: Taking the limit of this multiple variable function

1. ## Taking the limit of this multiple variable function

Hey, hope this is in the right topic

$\displaystyle f(x,y) = (sin(xy)-(xy))/sin(x^2+y^2)$

Thats whats troubling me, I just really dont know where to start and I cant work it out from the solutions.

Thanks guys.(And gals)

2. Originally Posted by Monkens
Hey, hope this is in the right topic

$\displaystyle f(x,y) = (sin(xy)-(xy))/sin(x^2+y^2)$

Thats whats troubling me, I just really dont know where to start and I cant work it out from the solutions.

Thanks guys.(And gals)
Which limit are you trying to take? The limit as what tends to what?

3. Originally Posted by Monkens
Hey, hope this is in the right topic

$\displaystyle f(x,y) = (sin(xy)-(xy))/sin(x^2+y^2)$

Thats whats troubling me, I just really dont know where to start and I cant work it out from the solutions.

Thanks guys.(And gals)
$\displaystyle x \to ? ,y \to ?$

4. I am very sorry, it's at the origin (0,0)

5. Originally Posted by Monkens
Hey, hope this is in the right topic

$\displaystyle f(x,y) = (sin(xy)-(xy))/sin(x^2+y^2)$

Thats whats troubling me, I just really dont know where to start and I cant work it out from the solutions.

Thanks guys.(And gals)
Let's prove the limit is 0. We know that $\displaystyle \sin(u)-u\sim-\frac{u^3}{6}$ when $\displaystyle u\to0$. Since $\displaystyle xy\to 0$ when $\displaystyle (x,y)\to(0,0)$, we have $\displaystyle \sin(xy)-xy\sim-\frac{(xy)^3}{6}$ when $\displaystyle (x,y)\to(0,0)$. Similarly, $\displaystyle \sin(x^2+y^2)\sim x^2+y^2$, so that $\displaystyle f(x,y)\sim-\frac{(xy)^3}{6(x^2+y^2)}$ when $\displaystyle (x,y)\to(0,0)$.

Now we can say for instance $\displaystyle \frac{|xy|^3}{x^2+y^2}\leq |xy^3|\frac{x^2}{x^2+y^2}\leq |xy^3|$ (since the ratio is less than 1) hence $\displaystyle \frac{(xy)^3}{6(x^2+y^2)}\to0$when $\displaystyle (x,y)\to(0,0)$. This concludes.

6. Thank you very much, that's really helpful I never thought of substituting u in then using the taylor polynomial for that.