Limit of Multivariable Function: x^2+y^2+2xy

fogel1497
Messages
10
Reaction score
0

Homework Statement



Find the limit of:
(x^2+y^2+2xy)/(x^2+y^2)

Homework Equations


x = r*cos(theta)
y= r*sin(theta)

The Attempt at a Solution



So what I did was change to polar coordinates. Then it simplifies to:

(r^2 + 2r^2cos(theta)sin(theta) )/r^2

Factoring out an r^2 from everything you get:
1 + 2cos(theta)sin(theta)

And the limit as r and theta go to zero would appear to be 1. However if I plug in numbers very close to zero on my calculator it tells me that the limit is 2. Little help?
 
Physics news on Phys.org
Your limit depends on theta. That means the limit is different as you approach (0,0) in different ways. E.g. if you approach along the x axis, x=t, y=0 (theta=0), you get 1. If you approach along a line x=t, y=t, (theta=pi/4) you get 2. What does this tell you about the existence of the limit?
 
It doesn't exist? So when i do that method if I find i have thetas in my answer then the limit does not exist?
 
If you get different answers for different values of theta, yes, the limit does not exist.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top