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Thread: limit question

  1. #1
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    Apr 2011
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    yet another limit question

    Hi,

    I have banged my head against this wall for hours:

    $\displaystyle lim_{z \to 1-i} \ \ x+i2x + iy = 1 + i$

    $\displaystyle
    z \in C, z=x+iy
    $

    I need to prove that the above is true, but without resorting to the theorem that limit of P(z) at a, where z and a are complex numbers, is P(a). nor do I want to use the theorem on limits of products or limit of two functions added together.

    I basically want to construct a Delta that would make the following statement true:

    $\displaystyle \forall\epsilon>0,\ \exists \delta>0 {, } \forall{z, 0<}\mid{z-(1-i)}\mid{ < }\delta{, }\mid (x + i2x + iy) - (1+i) \mid < \epsilon$

    I have tried a few different things:

    1. translated $\displaystyle x + i2x + iy$ to $\displaystyle z + i(z + \overline{z})$. I then calculated $\displaystyle 1/(z-(1-i))$, then multiplied that by the final expression that needs to be bound by $\displaystyle \epsilon$ in order to get a factor that could somehow by bound by a constant. This would have allowed me to work forward from the expression involving $\displaystyle \delta$ to the final expression bound by $\displaystyle \epsilon$.

    2. Tried to work directly with the x's and y's

    So far nothing has worked.

    Any help would be much appreciated.
    Last edited by avicenna; Apr 11th 2011 at 05:09 PM.
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  2. #2
    Senior Member
    Joined
    Feb 2008
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    Quote Originally Posted by avicenna View Post
    Hi,

    I have banged my head against this wall for hours:

    $\displaystyle lim_{z \to 1-i} \ \ x+i2x + iy = 1 + i$

    $\displaystyle
    z \in C, z=x+iy
    $

    I need to prove that the above is true, but without resorting to the theorem that limit of P(z) at a, where z and a are complex numbers, is P(a). nor do I want to use the theorem on limits of products or limit of two functions added together.

    I basically want to construct a Delta that would make the following statement true:

    $\displaystyle \forall\epsilon>0,\ \exists \delta>0 {, } \forall{z, 0<}\mid{z-(1-i)}\mid{ < }\delta{, }\mid (x + i2x + iy) - (1+i) \mid < \epsilon$

    I have tried a few different things:

    1. translated $\displaystyle x + i2x + iy$ to $\displaystyle z + i(z + \overline{z})$. I then calculated $\displaystyle 1/(z-(1-i))$, then multiplied that by the final expression that needs to be bound by $\displaystyle \epsilon$ in order to get a factor that could somehow by bound by a constant. This would have allowed me to work forward from the expression involving $\displaystyle \delta$ to the final expression bound by $\displaystyle \epsilon$.

    2. Tried to work directly with the x's and y's

    So far nothing has worked.

    Any help would be much appreciated.
    Let $\displaystyle \delta=\epsilon/4$. Then if

    $\displaystyle |x-1|,|y+1|\leq\sqrt{(x-1)^2+(y+1)^2}=|(x-1)+i(y+1)|=|z-(1-i)|<\delta$

    we have

    $\displaystyle |(x + i2x + iy) - (1+i)|=|(x-1)+i(2x+y-1)|\leq|x-1|+|2x+y-1|$

    $\displaystyle =|x-1|+|2x-2+y+1|\leq 3|x-1|+|y+1|< 4\delta=\epsilon$
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  3. #3
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    Quote Originally Posted by hatsoff View Post
    Let $\displaystyle \delta=\epsilon/4$. Then if

    $\displaystyle |x-1|,|y+1|\leq\sqrt{(x-1)^2+(y+1)^2}=|(x-1)+i(y+1)|=|z-(1-i)|<\delta$

    we have

    $\displaystyle |(x + i2x + iy) - (1+i)|=|(x-1)+i(2x+y-1)|\leq|x-1|+|2x+y-1|$

    $\displaystyle =|x-1|+|2x-2+y+1|\leq 3|x-1|+|y+1|< 4\delta=\epsilon$
    great! my hatsoff to you!
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  4. #4
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    Quote Originally Posted by hatsoff View Post
    Let $\displaystyle \delta=\epsilon/4$. Then if

    $\displaystyle |x-1|,|y+1|\leq\sqrt{(x-1)^2+(y+1)^2}=|(x-1)+i(y+1)|=|z-(1-i)|<\delta$

    we have

    $\displaystyle |(x + i2x + iy) - (1+i)|=|(x-1)+i(2x+y-1)|\leq|x-1|+|2x+y-1|$

    $\displaystyle =|x-1|+|2x-2+y+1|\leq 3|x-1|+|y+1|< 4\delta=\epsilon$
    one question hatsoff. Could you solve it if you first translate $\displaystyle x + i2x + iy$ to $\displaystyle z + i(z + \overline{z})$?
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