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Thread: two variable limit

  1. #1
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    two variable limit

    Hi! I am new and I can't solve this limit:

    limit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^6|+|y^4|)

    Can anyone give me a hand?
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  2. #2
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    Quote Originally Posted by venturozzaccio View Post
    Hi! I am new and I can't solve this limit:

    limit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^6|+|y^4|)

    Can anyone give me a hand?
    Try converting to polars:

    $\displaystyle \lim_{(x, y) \to (0,0)}\left(\frac{x^2y^4}{|x^6| + |y^4|}\right) = \lim_{r \to 0}\left[\frac{(r\cos{\theta})^2(r\sin{\theta})^4}{|(r\cos{ \theta})^6| + |(r\sin{\theta})^4|}\right]$

    $\displaystyle = \lim_{r \to 0}\left[\frac{r^6\cos^2{\theta}\sin^4{\theta}}{r^6|\cos^6{ \theta}| + r^4|\sin^4{\theta}|}\right]$

    $\displaystyle = \lim_{r \to 0}\left[\frac{r^2\cos^2{\theta}\sin^4{\theta}}{r^2|\cos^6{ \theta}| + |\sin^4{\theta}|}\right]$

    $\displaystyle = \frac{0^2\cos^2{\theta}\sin^4{\theta}}{0^2|\cos^6{ \theta}| + |\sin^4{\theta}|}$

    $\displaystyle = \frac{0}{|\sin^4{\theta}|}$

    $\displaystyle = 0$.
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  3. #3
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    sorry I make a mistake writing.

    the real one is :


    limit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^2|+|y^4|)
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Try converting to polars:

    $\displaystyle \lim_{(x, y) \to (0,0)}\left(\frac{x^2y^4}{|x^6| + |y^4|}\right) = \lim_{r \to 0}\left[\frac{(r\cos{\theta})^2(r\sin{\theta})^4}{|(r\cos{ \theta})^6| + |(r\sin{\theta})^4|}\right]$

    $\displaystyle = \lim_{r \to 0}\left[\frac{r^6\cos^2{\theta}\sin^4{\theta}}{r^6|\cos^6{ \theta}| + r^4|\sin^4{\theta}|}\right]$

    $\displaystyle = \lim_{r \to 0}\left[\frac{r^2\cos^2{\theta}\sin^4{\theta}}{r^2|\cos^6{ \theta}| + |\sin^4{\theta}|}\right]$

    $\displaystyle = \frac{0^2\cos^2{\theta}\sin^4{\theta}}{0^2|\cos^6{ \theta}| + |\sin^4{\theta}|}$

    $\displaystyle = \frac{0}{|\sin^4{\theta}|}$

    $\displaystyle = 0$.
    This is really nice
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  5. #5
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    yes but about the revised question?
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  6. #6
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    Use the exact same process. You will still be able to cancel enough $\displaystyle r$'s to be able to evaluate the limit.
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  7. #7
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    so the result is still 0 ?
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  8. #8
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    Quote Originally Posted by venturozzaccio View Post
    so the result is still 0 ?
    Correct.
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  9. #9
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    ok. but if the limit is


    limit x,y->(0,0) $\displaystyle xy^2 / (x^2+y^4) $

    then I have $\displaystyle 0cos()sen^2() /( cos^2+0^2cos^4())$

    And following the previous example this should be 0 again. But this time the limit doesn't exist. where I make a mistake?
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  10. #10
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    Quote Originally Posted by venturozzaccio View Post
    ok. but if the limit is


    limit x,y->(0,0) $\displaystyle xy^2 / (x^2+y^4) $

    then I have $\displaystyle 0cos()sen^2() /( cos^2+0^2cos^4())$

    And following the previous example this should be 0 again. But this time the limit doesn't exist. where I make a mistake?
    I get that the limit is $\displaystyle 0$...


    $\displaystyle \lim_{(x, y) \to (0, 0)}\frac{xy^2}{x^2 + y^4} = \lim_{r \to 0}\frac{r\cos{\theta}\,r^2\sin^2{\theta}}{r^2\cos^ 2{\theta} + r^4\sin^4{\theta}}$

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

    $\displaystyle = \lim_{r \to 0}\frac{r\cos{\theta}\sin^2{\theta}}{\cos^2{\theta } + r^2\sin^4{\theta}}$

    $\displaystyle = \frac{0}{\cos^2{\theta}}$

    $\displaystyle = 0$.
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  11. #11
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    I think it's not true. Just because if you set x=y the limit is 0. But if you set x=y^2 then the limit is 1/2. So limit change according the way you approach to (0,0) and for this it doesn't exist.
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