# Thread: Continuity in 2 variables.

1. ## Continuity in 2 variables.

Find the limit or show that the limit does not exist,

$\lim_{(x,y) \to (0,0)} \frac{xy \cos y}{3x^2+y^2}$

2. First we set...

$x= \rho\cdot \cos \theta$

$y= \rho\cdot \sin \theta$

... and imnmediately we obtain...

$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \cos y}{3 x^{2} + y^{2}} = \lim_{\rho \rightarrow 0} \frac{\rho^{2}\cdot \sin \theta\cdot \cos \theta \cdot \cos (\rho\cdot \sin \theta)}{\rho^{2}\cdot (3 \cos^{2} \theta + \sin^{2} \theta)} = \frac{\sin \theta \cos \theta}{1+2 \cos^{2} \theta}$

Since the limit depends form $\theta$ the limit itself doesn't exist...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
First we set...

$x= \rho\cdot \cos \theta$

$y= \rho\cdot \sin \theta$

... and imnmediately we obtain...

$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \cos y}{3 x^{2} + y^{2}} = \lim_{\rho \rightarrow 0} \frac{\rho^{2}\cdot \sin \theta\cdot \cos \theta \cdot \cos (\rho\cdot \sin \theta)}{\rho^{2}\cdot (3 \cos^{2} \theta + \sin^{2} \theta)} = \frac{\sin \theta \cos \theta}{1+2 \cos \theta}$

Since the limit depends form $\theta$ the limit itself doesn't exist...

Kind regards

$\chi$ $\sigma$
I find this very interesting. I haven't thought about taking limits of functions of two variables. Could you please justify the use of $p$ for me please?

4. The limit...

$\lim_{(x,y) \rightarrow (x_{0},y_{0})} f(x,y)$

... exists if and only if $f(x,y)$ tends to the same value $(x_{*},y_{*})$ no matter which is the 'traiectory' [straight line, zig zag, spiral, etc...] with which $(x,y)$ tends to $(x_{0},y_{0})$... this is the fundamental concept of limit of two [or more...] variables functions...

Kind regards

$\chi$ $\sigma$

5. Thanks allot, i just have another question,
Is the polar coordinate substition the best way to solve limits of 2 variables? or just one of the many? I have tried to use the formal definition of a limit, but I end up confusing myself..

For exmaple, I tried using the formal definition on that function, and dont go anywhere. Another one I was working on;
$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y}{\sqrt{ x^{2} + y^{2}}} = \lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}}$
So using the formal definition; $0 < \sqrt{ x^{2} + y^{2}}<\delta \mbox{\ then \ } |\frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}} - 0 < \epsilon$
Then i stated that $x \cdot y < x^2 + y^2 \mbox{ so } \frac{x\cdot y}{x^{2}+y^{2}} \leq 1$
But i wasn't sure where to take it form here, or even if this is on the right track...

6. Originally Posted by VonNemo19
I find this very interesting. I haven't thought about taking limits of functions of two variables. Could you please justify the use of $p$ for me please?
He introduced polar coordinates. Some might have seen

$
x = r \cos \theta,\; y = r \sin \theta
$

7. Originally Posted by Robb
Thanks allot, i just have another question,
Is the polar coordinate substition the best way to solve limits of 2 variables? or just one of the many? I have tried to use the formal definition of a limit, but I end up confusing myself..

For exmaple, I tried using the formal definition on that function, and dont go anywhere. Another one I was working on;
$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y}{\sqrt{ x^{2} + y^{2}}} = \lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}}$
So using the formal definition; $0 < \sqrt{ x^{2} + y^{2}}<\delta \mbox{\ then \ } |\frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}} - 0 < \epsilon$
Then i stated that $x \cdot y < x^2 + y^2 \mbox{ so } \frac{x\cdot y}{x^{2}+y^{2}} \leq 1$
But i wasn't sure where to take it form here, or even if this is on the right track...
Using polar coordinates we have...

$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y}{\sqrt{ x^{2} + y^{2}}} = \lim_{\rho \rightarrow 0} \frac{ \rho^{2}\cdot \sin \theta\cdot \cos \theta}{\rho} = 0$

Very easy! ...

Kind regards

$\chi$ $\sigma$

8. OMG.. thanks allot. I tried using the polar coordinates, but forgot the [/tex]\rho[/tex] from the denominator. So the condition is, that if $\rho$ is left in the function after substituting in the polar coordinates and simplifying then the limit exists?
The text i have is very light on using polar coordinates, it just has the sentenace (after using polar coordinates for a limit) that 'the behaviour as $\rho = \sqrt{x^2+y^2} \rightarrow 0$ depends on $\theta$ hence there is no limit'

Sorry for all the questions, i am just trying to get my head around this in a logical manner :P

9. Originally Posted by Robb
Thanks allot, i just have another question,
Is the polar coordinate substition the best way to solve limits of 2 variables? or just one of the many? I have tried to use the formal definition of a limit, but I end up confusing myself..

For exmaple, I tried using the formal definition on that function, and dont go anywhere. Another one I was working on;
$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y}{\sqrt{ x^{2} + y^{2}}} = \lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}}$
So using the formal definition; $0 < \sqrt{ x^{2} + y^{2}}<\delta \mbox{\ then \ } |\frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}} - 0 < \epsilon$
Then i stated that $x \cdot y < x^2 + y^2 \mbox{ so } \frac{x\cdot y}{x^{2}+y^{2}} \leq 1$
But i wasn't sure where to take it form here, or even if this is on the right track...
If you want to be a little more rigorous, (i.e. $\delta - \epsilon$) you could try using the inequalities

$
-\sqrt{x^2+y^2} \le x,y \le \sqrt{x^2+y^2}\;\;\Rightarrow\;\; -(x^2+y^2) \le x \,y \le x^2+y^2$

$\;\;\Rightarrow\;\;-\sqrt{x^2+y^2} \le \frac{x \,y}{\sqrt{x^2+y^2}} \le \sqrt{x^2+y^2}$

so

$\left| \frac{x \,y}{\sqrt{x^2+y^2}}\right| < \sqrt{x^2+y^2}$

Thus, if $\sqrt{x^2+y^2} < \epsilon$ then $\left| \frac{x \,y}{\sqrt{x^2+y^2}}\right| < \epsilon$.

10. Originally Posted by Robb
Find the limit or show that the limit does not exist,

$\lim_{(x,y) \to (0,0)} \frac{xy \cos y}{3x^2+y^2}$

You can examine a few paths and see if you keep getting the same result.
I like using y=mx, then let x head to zero.

OR try...
(1) y=o and let x tend to zero, the limit along both parts of the x-axis as we head to the origin is 0, since you have 0 over $3x^2$.

And for a second path, just let x=y, which is just calculus one now...
(2) In this case the limit is

$\lim_{x\to 0}{x^2\cos x\over 4x^2}={1\over 4}$

SINCE two different paths produce two different limits, the limit does not exist as we approach the origin.