# Taking the limit of this multiple variable function

• May 25th 2010, 03:34 PM
Monkens
Taking the limit of this multiple variable function
Hey, hope this is in the right topic

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

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)
• May 25th 2010, 03:40 PM
Mush
Quote:

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

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

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?
• May 25th 2010, 03:44 PM
pickslides
Quote:

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

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

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)

$
x \to ? ,y \to ?
$
• May 26th 2010, 06:23 AM
Monkens
I am very sorry, it's at the origin (0,0)
• May 26th 2010, 07:40 AM
Laurent
Quote:

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

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

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 $\sin(u)-u\sim-\frac{u^3}{6}$ when $u\to0$. Since $xy\to 0$ when $(x,y)\to(0,0)$, we have $\sin(xy)-xy\sim-\frac{(xy)^3}{6}$ when $(x,y)\to(0,0)$. Similarly, $\sin(x^2+y^2)\sim x^2+y^2$, so that $f(x,y)\sim-\frac{(xy)^3}{6(x^2+y^2)}$ when $(x,y)\to(0,0)$.

Now we can say for instance $\frac{|xy|^3}{x^2+y^2}\leq |xy^3|\frac{x^2}{x^2+y^2}\leq |xy^3|$ (since the ratio is less than 1) hence $\frac{(xy)^3}{6(x^2+y^2)}\to0$when $(x,y)\to(0,0)$. This concludes.
• May 26th 2010, 09:16 AM
Monkens
Thank you very much, that's really helpful :) I never thought of substituting u in then using the taylor polynomial for that.