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Math Help - Another limit question

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Another limit question

    I know this is very intuitive but I just wanted to make sure.

    Consider

    \lim_{(x,y)\to(a,b)}f(x,y)

    Now let \lim_{(x,y)\to(a,b)}f(x,y)=L\text{ Along }P_1

    But suppose

    \lim_{(x,y)\to(a,b)}f(x,y)\text{ Does not exist along }P_2

    Then obviously

    \lim_{(x,y)\to(a,b)}f(x,y)\text{ Does not exist}

    Right?


    Example

    \lim_{(x,y)\to(0,0)}\frac{3x^2\sqrt{y}}{x^4+y^2}

    Along the y-axis we have

    \lim_{(0,y)\to(0,0)}\frac{3(0)^2\sqrt{y}}{0^4+y^2}  =0

    Now along the parabola y=x^2

    We get

    \lim_{(x,x^2)\to(0,0)}\frac{3x^3|x|}{x^4+x^4}=\fra  c{3}{2}\lim_{(x,x^2)\to(0,0)}\frac{|x|}{x}

    Which does not exist.

    So then just to make sure, I can say that since

    \lim_{(0,y)\to(0,0)}\frac{3(0)^3\sqrt{y}}{0^4+y^2}  \ne\lim_{(x,x^2)\to(0,0)}\frac{3x^3\sqrt{x^2}}{x^4  +x^4}

    That

    \lim_{(x,y)\to(0,0)}\frac{3x^3\sqrt{y}}{x^4+y^2} does not exist right?

    Thanks in advance,
    Mathstud
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by topsquark View Post
    Your argument looks correct, but you might want to check this result.

    -Dan
    Are you saying it isn't zero?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Are you saying it isn't zero?
    (Ahem!) I read the numerator as adding, not multiplying.

    My apologies!

    -Dan
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  4. #4
    Eater of Worlds
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    If I am understanding you correctly, you are right, the limit does not exist because it approaches different values along different paths.

    Here is an example:

    \lim_{(x,y)\to (0,0)}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)^{2}

    The domain of said function consists of all points in the xy-plane for the point (0,0). To show that the limit approaches (0,0) does not exist, consider approaching (0,0) along two different paths. Along the x-axis, every point has the form (x,0) and the limit is 1.

    If (x,y) approaches (0,0) along y=x, we get 0

    This means that in any open disc centered at (0,0) there are points (x,y) where the function takes on the value 1, and other places where it has value 0.

    Therefore, it does not have a limit as (x,y)\rightarrow (0,0)

    Now, we can conclude that the limit does not exist because we found two different values for two different approaches. If two approaches had given the same limit, we still could not have concluded the limit exists.

    To form this conclusion, we must show the limit is the same along all possible approaches.

    I am sure you already knew this.

    Here is something else:

    \lim_{(x,y)\to (0,0)}\frac{5x^{2}y}{x^{2}+y^{2}}

    In this case, the limits of the num. and den. are both 0, so we can not determine the existence of a limit by taking the limits of the num. and den. separately and dividing.

    Note that |y|\leq\sqrt{x^{2}+y^{2}}

    and

    \frac{x^{2}}{x^{2}+y^{2}}\leq 1

    Then, in a neighborhood, {\delta}, about (0,0), you have

    0<\sqrt{x^{2}+y^{2}}<{\delta}, and it

    follows that, for (x,y)\neq (0,0)

    |f(x,y)-0|=\left|\frac{5x^{2}y}{x^{2}+y^{2}}\right|

    =5|y|\left(\frac{x^{2}}{x^{2}+y^{2}}\right)

    \leq 5|y|

    \leq 5\sqrt{x^{2}+y^{2}}

    \leq 5{\delta}

    Therefore, hence, we can choose {\delta}=\frac{\epsilon}{5}

    and conclude that \lim_{(x,y)\to (0,0)}\frac{5x^{2}y}{x^{2}+y^{2}}=0

    I hope you liked this little tutorial, though, you are probably already familiar.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    If I am understanding you correctly, you are right, the limit does not exist because it approaches different values along different paths.

    Here is an example:

    \lim_{(x,y)\to (0,0)}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)^{2}

    The domain of said function consists of all points in the xy-plane for the point (0,0). To show that the limit approaches (0,0) does not exist, consider approaching (0,0) along two different paths. Along the x-axis, every point has the form (x,0) and the limit is 1.

    If (x,y) approaches (0,0) along y=x, we get 0

    This means that in any open disc centered at (0,0) there are points (x,y) where the function takes on the value 1, and other places where it has value 0.

    Therefore, it does not have a limit as (x,y)\rightarrow (0,0)

    Now, we can conclude that the limit does not exist because we found two different values for two different approaches. If two approaches had given the same limit, we still could not have concluded the limit exists.

    To form this conclusion, we must show the limit is the same along all possible approaches.

    I am sure you already knew this.

    Here is something else:

    \lim_{(x,y)\to (0,0)}\frac{5x^{2}y}{x^{2}+y^{2}}

    In this case, the limits of the num. and den. are both 0, so we can not determine the existence of a limit by taking the limits of the num. and den. separately and dividing.

    Note that |y|\leq\sqrt{x^{2}+y^{2}}

    and

    \frac{x^{2}}{x^{2}+y^{2}}\leq 1

    Then, in a neighborhood, {\delta}, about (0,0), you have

    0<\sqrt{x^{2}+y^{2}}<{\delta}, and it

    follows that, for (x,y)\neq (0,0)

    |f(x,y)-0|=\left|\frac{5x^{2}y}{x^{2}+y^{2}}\right|

    =5|y|\left(\frac{x^{2}}{x^{2}+y^{2}}\right)

    \leq 5|y|

    \leq 5\sqrt{x^{2}+y^{2}}

    \leq 5{\delta}

    Therefore, hence, we can choose {\delta}=\frac{\epsilon}{5}

    and conclude that \lim_{(x,y)\to (0,0)}\frac{5x^{2}y}{x^{2}+y^{2}}=0

    I hope you liked this little tutorial, though, you are probably already familiar.
    Indeed I was! But repetition never hurt anything ....unless its Shakespeare
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