Results 1 to 10 of 10

Math Help - Continuity in 2 variables.

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    133

    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}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by chisigma View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2009
    Posts
    133
    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...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    Quote Originally Posted by VonNemo19 View Post
    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

     <br />
x = r \cos \theta,\; y = r \sin \theta<br />
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by Robb View Post
    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Mar 2009
    Posts
    133
    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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    Quote Originally Posted by Robb View Post
    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

     <br />
-\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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by Robb View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Continuity of a function of 2 variables
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 10th 2010, 02:11 PM
  2. Functions of two variables continuity question
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 15th 2010, 11:28 AM
  3. Continuity of a 2 variables function in R2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 16th 2009, 05:17 AM
  4. Continuity of a 2 variables function
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 15th 2009, 08:56 AM
  5. Function in two variables - continuity
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 21st 2009, 10:45 PM

Search Tags


/mathhelpforum @mathhelpforum