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Math Help - please enter and help me>>>thanks

  1. #1
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    please enter and help me>>>thanks

    find the limits by referring details when you solve them when (x,y)goes to (0,0) :

    1_ lim (x^2 + y^2 ) sin ( 1\ (x
    ^2 + y^2))(i'm trying but i don't know )

    2- lim e ^ -1 over (sqr root for
    x^2 + y^2)

    3- lim
    { e ^ -1 over (sqr root for x^2 + y^2) } over (sqr root for x^2 + y^2)

    4- lim 1-
    ( x^2 + y^2) \ ( x^2 + y^2)

    5- lim y ln
    ( x^2 + y^2)

    when i solve the limit #4 first i substitute (0,0) the answer 1\0 i know that the limit doesn't exist but i don't know how i can show that ???

    in #5 when i translate xy-coordinate to r,theta i can't complete


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  2. #2
    Moo
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    Hello,

    Have you tried converting all rectangular coordinates into polar coordinates ??

    x=r \cos \theta
    y=r \sin \theta

    x^2+y^2=r^2

    (x,y) \to (0,0) \Leftrightarrow r \to 0
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    Have you tried converting all rectangular coordinates into polar coordinates ??

    x=r \cos \theta
    y=r \sin \theta

    x^2+y^2=r^2

    (x,y) \to (0,0) \Leftrightarrow r \to 0
    yes, but i don't know why we can't use H˘pital's rule when we do parametrization for x and y ??? (we get one variable is t )


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  4. #4
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    Quote Originally Posted by flower3
    1_ lim (x^2 + y^2 ) sin ( 1\ (x[/FONT][/FONT][FONT=Tahoma][FONT=Tahoma] ^2 + y^2))(i'm trying but i don't know )
    For this one note that,
    0\leq (x^2+y^2) \sin \frac{1}{x^2+y^2} \leq (x^2+y^2)
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  5. #5
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    please help me>>>>>>i don't know how

    please help me>>>>>>i don't know how
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  6. #6
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    Reply

    I'll help you on the first one because your notation is a little off on the rest:

    (x^2+y^2) sin(1/(x^2+y^2))


    Let x^2+y^2=r^2 through parametric transformation:

    (r^2) sin(1/(r^2))

    Now if we know that r=0 (because x^2+y^2=0), then we know we have:

    0* sin(1/(r^2))

    Regardless of what the sin value produces, you know it is within the limits of the sine function solution (-1 through +1), and because zero times any constant, even in this case zero, should the sine value equal zero, would still be zero^2=0.

    Can you get it from here?
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