Results 1 to 4 of 4

Math Help - limit question

  1. #1
    Newbie
    Joined
    Apr 2011
    Posts
    11

    yet another limit question

    Hi,

    I have banged my head against this wall for hours:

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

    <br />
z \in C, z=x+iy<br />

    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:

    \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 x + i2x + iy to z + i(z + \overline{z}). I then calculated 1/(z-(1-i)), then multiplied that by the final expression that needs to be bound by \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 \delta to the final expression bound by \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; April 11th 2011 at 05:09 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by avicenna View Post
    Hi,

    I have banged my head against this wall for hours:

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

    <br />
z \in C, z=x+iy<br />

    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:

    \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 x + i2x + iy to z + i(z + \overline{z}). I then calculated 1/(z-(1-i)), then multiplied that by the final expression that needs to be bound by \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 \delta to the final expression bound by \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 \delta=\epsilon/4. Then if

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

    we have

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

    =|x-1|+|2x-2+y+1|\leq 3|x-1|+|y+1|< 4\delta=\epsilon
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2011
    Posts
    11
    Quote Originally Posted by hatsoff View Post
    Let \delta=\epsilon/4. Then if

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

    we have

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

    =|x-1|+|2x-2+y+1|\leq 3|x-1|+|y+1|< 4\delta=\epsilon
    great! my hatsoff to you!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2011
    Posts
    11
    Quote Originally Posted by hatsoff View Post
    Let \delta=\epsilon/4. Then if

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

    we have

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

    =|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 x + i2x + iy to z + i(z + \overline{z})?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: November 7th 2011, 03:27 PM
  2. limit question
    Posted in the Differential Geometry Forum
    Replies: 10
    Last Post: February 10th 2011, 10:46 AM
  3. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  4. Limit question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 26th 2009, 03:55 AM
  5. Limit Question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 17th 2009, 10:25 AM

Search Tags


/mathhelpforum @mathhelpforum