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Math Help - Function, using two-path test, finding limits

  1. #1
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    Function, using two-path test, finding limits

    f(x,y) = \frac{x}<br />
{{1 - {y^2}}},y \ne  \pm 1<br />

    a) Using the paths x=0 and y=x+1, show by the two-path test that the limit, {\lim _{(x,y) \to (0,1)}}f(x,y)<br />
does not exist.


    b) Find the limit, {\lim _{(x,y) \to (1,0)}}f(x,y)<br />
, if it exists. IF it does not, explain why not.

    c) Is it possible to define a new function g(x,y) that is defined and continuous for all (x,y) in R^2, and such that g(x,y)=f(x,y) for all (x,y) in the domain of f? If so, find such a function, If not, explain why not.



    I'm not sure where to start for a)

    for b) i think its {\lim _{(x,y) \to (0,1)}}\frac{x}<br />
{{1 - {y^2}}} = \frac{1}<br />
{{1 - {{(0)}^2}}} = 1<br />

    and I also don't know where to start for c)
    Last edited by genlovesmusic09; April 29th 2010 at 06:00 PM.
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  2. #2
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    Quote Originally Posted by genlovesmusic09 View Post
    f(x,y) = \frac{x}<br />
{{1 - {y^2}}},y \ne  \pm 1<br />

    a) Using the paths x=0 and y=x+1, show by the two-path test that the limit, {\lim _{(x,y) \to (0,1)}}f(x,y)<br />
does not exist.


    b) Find the limit, {\lim _{(x,y) \to (0,1)}}f(x,y)<br />
, if it exists. IF it does not, explain why not.
    Is this the same "f(x,y)" as in (a)? If so, doesn't (a) already ask you to show that this limit does not exist? Are you sure you have copied these problems correctly? Perhaps this was the limit as (x, y) goes to (0, -1)?

    c) Is it possible to define a new function g(x,y) that is defined and continuous for all (x,y) in R^2, and such that (9x,y)=f(x,y) for all (x,y) in the domain of f? If so, find such a function, If not, explain why not.

    I'm not sure where to start for a)
    I recommend you start by doing what you were told to do, " Using the paths x=0 and y=x+1"!
    On x=0, f(x,y)= \frac{0}{1- y^2}= 0, as long as y is not 1 or -1, for all x so the limit, as y goes to 1, is 0.

    On y= x+ 1, x= y- 1 so f(x,y)= \frac{y- 1}{1- y^2}= \frac{y-1}{(1-y)(1+y)} and \lim_{y\to 1}\frac{y-1}{1- y^2}= \lim_{y\to 1}\frac{1}{1+y}= \frac{1}{2}

    for b) i think its {\lim _{(x,y) \to (0,1)}}\frac{x}<br />
{{1 - {y^2}}} = \frac{1}<br />
{{1 - {{(0)}^2}}} = 1<br />
    What?? You've set x= 1 and y= 0! (0, 1) means x= 0 and y= 1.

    and I also don't know where to start for c)
    I assume the "9" is a typo. Do you mean g(x,y)= f(x,y)? Since f(x,y) is defined for all (x,y) except (x, 1) and (x, -1), such a function would have to have the same limit as (x, y) goes to (x, 1) or as (x, y) goes to (x, -1), for any x. And that is possible only if that limit exist.
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  3. #3
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    Sorry, I've fixed the mistakes for all the problems, b) is really lin from (1,) and I fixed part of c)

    Sorry again!


    hopefully this makes more sense...


    also for the part a) I've thought about it more and the two paths yest i've used it that when x=o the limit = 0 and when y=x+1 the limit is -1/2 so it does not exist because the limits are different

    I'm still lost for but c though...
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