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Math Help - Problem with a limit's existence...

  1. #1
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    Problem with a limit's existence...

    Heya guys. I have to study the continuity of the following function at (0,0):

    f(x,y) =\begin{cases}<br />
	\frac{x^3y}{x^4+y^2}  & (x,y)\neq(0,0) \\<br />
	0  & (x,y)=(0,0)<br />
\end{cases}

    Thus I need to check whether

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

    exists or not.

    I converted to polar coordinates:

    	\lim_{r\to 0} \frac{r^4 cos(\theta)^3 sin(\theta)}{(r cos(\theta))^4+(r sin(\theta))^2} = \lim_{r\to 0} \frac{r^2 cos(\theta)^3 sin(\theta)}{r^2 cos(\theta)^4+sin(\theta)^2} = \frac{0}{sin(\theta)^2}

    Thus the limit seems to be 0 except where

    sin(\theta)^2=0 \implies sin(\theta)=0 \implies \theta = 0 \vee \theta = \pi

    If \theta = 0 \vee \theta = \pi, then

    \lim_{r\to 0} \frac{r^4 cos(\theta)^3 sin(\theta)}{(r cos(\theta))^4+(r sin(\theta))^2} = \frac{0}{0}

    I then used L'Hopital's rule:

    \lim_{r\to 0} \frac{r^2 cos(\theta)^3 sin(\theta)}{r^2 cos(\theta)^4+sin(\theta)^2} = \lim_{r\to 0} \frac{cos(\theta)^3 sin(\theta)}{cos(\theta)^4} = 0

    since sin(\theta)=0.

    However, by using such paths as y=x and y=2x, the limit equals 1 and 2, respectively. What's going on that I don't understand here?

    Thanks!
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  2. #2
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    Quote Originally Posted by FatherChaos View Post
    I have to study the continuity of the following function at (0,0):

    f(x,y) =\begin{cases}<br />
    \frac{x^3y}{x^4+y^2}  & (x,y)\neq(0,0) \\<br />
    0  & (x,y)=(0,0)<br />
\end{cases}

    ... [snip] ...

    However, by using such paths as \color{magenta}y=x and \color{magenta}y=2x, the limit equals 1 and 2, respectively.
    That last statement is not true. If y=kx (with k ≠ 0) then f(x,y) = \frac{kx^4}{x^4+y^2} = \frac{kx^2}{x^2+k^2}, and so \lim_{(x,y)\to(0,0)}f(x,y) = 0.

    Hint: write f(x,y) as x\cdot\frac{x^2y}{x^4+y^2}, and show that \frac{x^2y}{x^4+y^2}\leqslant\frac12.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Cool

    Also see the following graphic.

    ..........The numerator, Absolute value of the numerator, The denominator

    Codes for Mathematica

    Code:
    Plot3D[{Abs[x^3y],x^3y,x^4+y^2},{x,-1/2,1/2},{y,-1/2,1/2},PlotStyle->{{Red},{Yellow},{Blue}}]
    Can you see that the (absolute value of) numerator is closer to 0 than the denominator?
    So this implies that the limit must be 0 at (0,0).
    To prove this mathematically, you can lead the direction Opalg points.
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  4. #4
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    Quote Originally Posted by Opalg View Post
    That last statement is not true. If y=kx (with k ≠ 0) then f(x,y) = \frac{kx^4}{x^4+y^2} = \frac{kx^2}{x^2+k^2}, and so \lim_{(x,y)\to(0,0)}f(x,y) = 0.

    Hint: write f(x,y) as x\cdot\frac{x^2y}{x^4+y^2}, and show that \frac{x^2y}{x^4+y^2}\leqslant\frac12.
    Thanks a lot. Somehow my teacher's corrected version of this exercise was flat-out wrong. I still have a question though. What is the point of showing that \frac{x^2y}{x^4+y^2}\leqslant\frac12, exactly? I do not get the big picture.
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  5. #5
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    Quote Originally Posted by FatherChaos View Post
    What is the point of showing that \frac{x^2y}{x^4+y^2}\leqslant\frac12, exactly? I do not get the big picture.
    The point is that it then follows that |f(x,y)|= \left|x\,\frac{x^2y}{x^4+y^2}\right|\leqslant\tfra  c12|x|, and so \lim_{(x,y)\to(0,0)}f(x,y) = 0.
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  6. #6
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    Quote Originally Posted by Opalg View Post
    The point is that it then follows that |f(x,y)|= \left|x\,\frac{x^2y}{x^4+y^2}\right|\leqslant\tfra  c12|x|, and so \lim_{(x,y)\to(0,0)}f(x,y) = 0.
    Wow, you're right! And I came to the conclusion that \frac{x^2y}{x^4+y^2}\leqslant\tfrac12 \implies (x^2-y)^2\geqslant 0, which is obviously true. Thus, the limit exists. Thanks a bunch ! Also, one last question. Why is it that I always see those absolute values when you use the squeeze theorem? Are they really needed?
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