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  1. #1
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    limit

    prove that:

     lim_{(x,y)\to (1,1)}\frac{2+x}{y} = 3

    Again i know the definition but i have no idea how to get δ
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  2. #2
    Super Member girdav's Avatar
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    We use the definition "with \epsilon and \delta" to obtain the rules about limits and sum or product.
    In cases like that, you can use these results instead of the " \epsilon".
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  3. #3
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    Quote Originally Posted by alexandros View Post
    prove that:

     lim_{(x,y)\to (1,1)}\frac{2+x}{y} = 3

    Again i know the definition but i have no idea how to get δ
    Are you required to use and " \epsilon, \delta" proof?

    You can, for example, replace x with x'= x- 1 so x= x'+1 and y with y'= y- 1so y= y'+1. Then, the problem becomes to prove that \frac{2+ x'+1}{y'+1}= \frac{3+ x'}{y'+1} goes to 3 as x' and y' go to 0. Now change to polar coordinates: x'= r cos(\theta) and y'= r sin(\theta). Now we have \frac{r cos(\theta)+ 3}{r sin(\theta)+ 1}. The reason I "shifted" the point to (0, 0) is that r alone measures the distance to the origin. Now, as r goes to 0, both r cos(\theta) and r sin(\theta) go to 0 as r goes to 0, no matter what \theta is and so the limit is 3.

    And, now, I think it would not be too difficult to get a \delta, \epsilon proof from that. Given any epsilon, we want \delta so that if \sqrt{(x-1)^2+ (y-1)^2}= r< \delta then |\frac{2+ x}{y}- 3|= |\frac{rcos(\theta)+ 3}{r sin(\theta)+ 1}- 3| = \frac{r cos(\theta)- 3r sin(\theta)}{rsin(\theta)+ 1}|= r|\frac{cos(\theta)- 3sin(\theta)}{r sin(\theta)+ 1}|< \epsilon

    Find an upper bound on |\frac{cos(\theta)- 3sin(\theta)}{r sin(\theta)+ 1}| and take \delta to be less than \epsilon divided by that upper bound.
    Last edited by HallsofIvy; January 25th 2010 at 06:41 AM.
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  4. #4
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    Why ,do we have to make so many substitutions ,can we not in straight forward manner using the ε-δ definition find the limit?

    The definition i was given is:

    given ε>0 find a δ>0 such that :

    if 0<||(x,y)-(1,1)||_{Eu}<\delta ,then |\frac{2+x}{y}-3|<\epsilon
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