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)

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- May 25th 2010, 02:34 PMMonkensTaking 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) - May 25th 2010, 02:40 PMMush
- May 25th 2010, 02:44 PMpickslides
- May 26th 2010, 05:23 AMMonkens
I am very sorry, it's at the origin (0,0)

- May 26th 2010, 06:40 AMLaurent
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. - May 26th 2010, 08:16 AMMonkens
Thank you very much, that's really helpful :) I never thought of substituting u in then using the taylor polynomial for that.