# Thread: limit question

1. ## yet another limit question

Hi,

I have banged my head against this wall for hours:

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

$
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:

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

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.

2. Originally Posted by avicenna
Hi,

I have banged my head against this wall for hours:

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

$
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:

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

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

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

$=|x-1|+|2x-2+y+1|\leq 3|x-1|+|y+1|< 4\delta=\epsilon$

3. Originally Posted by hatsoff
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!

4. Originally Posted by hatsoff
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 $z + i(z + \overline{z})$?