# Thread: Limit of a 2-variable function

1. ## Limit of a 2-variable function

Hello everyone!

I am trying to find a limit for the 2-variavle function $f(x,y)=\frac{(x+y)^2}{x^2+y^2}$.

So right before changing to polar coordinates, I get something like:
$L=1+\lim_{(x,y) \rightarrow (+\infty ,+\infty)} \, \frac{2xy}{x^2+y^2}$ $=1+\lim_{r \rightarrow +\infty} \, \frac{2r^2 \sin \theta \cos \theta}{r^2}$ and thus $L$ is bounded by 0 and 2: $0.

(1) Does this mean that the limit exists?
(2) Can we conclude that the numerator and the denomenator are of the same "order"?
(because the ratio of the numerator and the denominator is a constant)

Any help is appreciated!

2. Easier: the limits along $y=x$ and $y=-x$ are different.

Regards.

Fernando Revilla

3. Okay. I may need to add something. Since, x and y are approching $+\infty$ ten the angle $\theta$ is between 0 and 90 degrees so $1.
You're right, it's much easier, but my approach was to get a limit that is not zero or $\infty$ to determine wether the numerator and denominator are of same order eg. $x^3+\log x$ and $4x^3-\log 2x -10$ are of same order...

4. I understand. But even if you consider $0\leq \theta \leq \pi/2$ we have:

$f(\rho\cos \theta,\rho \sin \theta)=1+\sin 2\theta$

so, the limit varies according to $\theta$

Regards.

Fernando Revilla

5. If you're going to use polars, note that $\displaystyle \frac{(x + y)^2}{x^2 + y^2} = \frac{x^2 + y^2 + 2xy}{x^2 + y^2}$.

Using the change of coordinates $\displaystyle x = r\cos{\theta}, y = r\sin{\theta}, x^2 + y^2 = r^2$ we find

$\displaystyle \frac{x^2 + y^2 + 2xy}{x^2 + y^2} = \frac{r^2 + 2r^2\sin{\theta}\cos{\theta}}{r^2} = 1 + 2\sin{\theta}\cos{\theta}$.

Notice that this does not depend on $\displaystyle r$, so making $\displaystyle r \to \infty$ won't make any difference.

Since the limit will change over different paths (depending on the value of $\displaystyle \theta$) we can conclude that the limit does not exist.