Find the limit or show that the limit does not exist,
$\displaystyle \lim_{(x,y) \to (0,0)} \frac{xy \cos y}{3x^2+y^2}$
First we set...
$\displaystyle x= \rho\cdot \cos \theta$
$\displaystyle y= \rho\cdot \sin \theta$
... and imnmediately we obtain...
$\displaystyle \lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \cos y}{3 x^{2} + y^{2}} = \lim_{\rho \rightarrow 0} \frac{\rho^{2}\cdot \sin \theta\cdot \cos \theta \cdot \cos (\rho\cdot \sin \theta)}{\rho^{2}\cdot (3 \cos^{2} \theta + \sin^{2} \theta)} = \frac{\sin \theta \cos \theta}{1+2 \cos^{2} \theta}$
Since the limit depends form $\displaystyle \theta$ the limit itself doesn't exist...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
The limit...
$\displaystyle \lim_{(x,y) \rightarrow (x_{0},y_{0})} f(x,y)$
... exists if and only if $\displaystyle f(x,y)$ tends to the same value $\displaystyle (x_{*},y_{*})$ no matter which is the 'traiectory' [straight line, zig zag, spiral, etc...] with which $\displaystyle (x,y)$ tends to $\displaystyle (x_{0},y_{0})$... this is the fundamental concept of limit of two [or more...] variables functions...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
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;
$\displaystyle \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; $\displaystyle 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 $\displaystyle x \cdot y < x^2 + y^2 \mbox{ so } \frac{x\cdot y}{x^{2}+y^{2}} \leq 1$
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...
$\displaystyle \lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y}{\sqrt{ x^{2} + y^{2}}} = \lim_{\rho \rightarrow 0} \frac{ \rho^{2}\cdot \sin \theta\cdot \cos \theta}{\rho} = 0$
Very easy! ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
OMG.. thanks allot. I tried using the polar coordinates, but forgot the [/tex]\rho[/tex] from the denominator. So the condition is, that if $\displaystyle \rho$ 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 $\displaystyle \rho = \sqrt{x^2+y^2} \rightarrow 0 $ depends on $\displaystyle \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
If you want to be a little more rigorous, (i.e. $\displaystyle \delta - \epsilon $) you could try using the inequalities
$\displaystyle
-\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 $
$\displaystyle \;\;\Rightarrow\;\;-\sqrt{x^2+y^2} \le \frac{x \,y}{\sqrt{x^2+y^2}} \le \sqrt{x^2+y^2}$
so
$\displaystyle \left| \frac{x \,y}{\sqrt{x^2+y^2}}\right| < \sqrt{x^2+y^2}$
Thus, if $\displaystyle \sqrt{x^2+y^2} < \epsilon $ then $\displaystyle \left| \frac{x \,y}{\sqrt{x^2+y^2}}\right| < \epsilon$.
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 $\displaystyle 3x^2$.
And for a second path, just let x=y, which is just calculus one now...
(2) In this case the limit is
$\displaystyle \lim_{x\to 0}{x^2\cos x\over 4x^2}={1\over 4}$
SINCE two different paths produce two different limits, the limit does not exist as we approach the origin.