Finding the Limit of a Complex Function

[Bashyboy]

Hello everyone,

How do I find the limit of a complex function from the definition of a limit? For instance, consider the limit

$lim_{z \rightarrow -3} (5z+4i)$.

Would I simply conjecture that $5z + 4i$ approaches $5(-3) + 4i$ as $z \rightarrow -3$; and then use the definition of a limit to justify this?

 

[Mark44]

Hello everyone,

How do I find the limit of a complex function from the definition of a limit? For instance, consider the limit

$lim_{z \rightarrow -3} (5z+4i)$.

Would I simply conjecture that $5z + 4i$ approaches $5(-3) + 4i$ as $z \rightarrow -3$; and then use the definition of a limit to justify this?

Yout don't need to use the definition of the limit to find the limit value in this case. Just plug -3 into get the limit, which is -15 + 4i. That's pretty easy; proving it will take a bit more work.

 

[Bashyboy]

I have another question. In the definition of a limit, why aren't the quantifiers reversed, so that the definition of a limit read $\exists \epsilon >0$ $\forall \delta >0...$?

 

[GFauxPas]

\forall \epsilon > 0: \left \vert {f\left({x}\right) - L}\right \vert < \epsilon means "no matter how small the distance between f(x) and L is", i.e., "for any degree of closeness you can think of, no matter how small".

\exists \delta > 0: 0 < \vert x - c \vert < \delta means that I can guarantee you there's a way to get that close. The way you get that close is by making x very close to c, but not actually at c.

 

[Mark44]

I've deleted my earlier post. I misread what Bashyboy wrote, mistakenly thinking that all he did was reverse the order of the quantifiers.

 

[Bashyboy]

Hmm...It still seems as though it should be reversed. It seems that, no matter how close $z$ gets to $z_0$, $f(z)$ should also be close to $L$; that is, for every distance $\delta$ I chose, there should exist a corresponding distance $\epsilon$.

 

[Mark44]

Hmm...It still seems as though it should be reversed. It seems that, no matter how close $z$ gets to $z_0$, $f(z)$ should also be close to $L$; that is, for every distance $\delta$ I chose, there should exist a corresponding distance $\epsilon$.

That's not the way to think about it. The goal is not to show that some $\epsilon$ exists - it's to show that it can be made arbitrarily small.

Think of the whole limit business as a dialog between you (the prover) and a skeptic you're trying to convince.

He says, "OK, can you make f(z) within 0.1 of L?"
You say, "Sure, take $\delta$ to be 0.05. Then for every z within 0.05 of z₀, f(z) is within 0.1 of L. Are you convinced?"
He says, "Not yet. Can you get f(z) within 0.01 of L?"
You say, "Sure, take $\delta$ to be 0.0025. Then for every z within 0.025 of z₀, f(z) is within 0.01 of L. Are you convinced now?"
.
.
.
Finally, he gives up after realizing that no matter what he chooses for $\epsilon$ (this is the $\forall \epsilon$ part) you can find a $\delta$ (the $\exists \delta$ part).

 

[Bashyboy]

Okay, so I have a another complex limit I am dealing with. Here is my work:

$\lim\limits_{z \rightarrow (2+4i)} f(z) = 5i$, where $f(z) = 5i$. Let $\epsilon >0$ be arbitrary.

$|f(z) - 5i| < \epsilon$

$|5i - 5i| < \epsilon$

$|5i - 5i + (z-z) + (2-2)| < \epsilon$

$|(5i-z-2) + (z-5i+2)| < \epsilon$

Using the triangle inequality,

$|5i-z-2| + |z-5i-2| < \epsilon$

$|-1 \cdot (z + 2 - 5i)|| + |z-5i-2| < \epsilon$

$|z + 2 - 5i| + |z + 2 - 5i| < \epsilon$

$|z + 2 - 5i| < \frac{\epsilon}{2}$

$|z - 2 + 4 - 4i - i| < \frac{\epsilon}{2}$

$|(z - 2 - 4i) + (4-i)| < \frac{\epsilon}{2}$

$|z - (2 + 4i)| +|4-i| < \frac{\epsilon}{2}$

$|z - (2 + 4i)| < \frac{\epsilon}{2} - \sqrt{17}$

If I let $\delta = \frac{\epsilon}{2} - \sqrt{17}$, isn't possible that it could be negative?

 

[Mark44]

Okay, so I have a another complex limit I am dealing with. Here is my work:

$\lim\limits_{z \rightarrow (2+4i)} f(z) = 5i$, where $f(z) = 5i$. Let $\epsilon >0$ be arbitrary.

$|f(z) - 5i| < \epsilon$

$|5i - 5i| < \epsilon$

Since f(z) is constant, no matter what $\epsilon$ anyone chooses, you have complete freedom with $\delta$ - any value of $\delta$ will work. None of the work below is necessary.

$|5i - 5i + (z-z) + (2-2)| < \epsilon$

$|(5i-z-2) + (z-5i+2)| < \epsilon$

Using the triangle inequality,

$|5i-z-2| + |z-5i-2| < \epsilon$

$|-1 \cdot (z + 2 - 5i)|| + |z-5i-2| < \epsilon$

$|z + 2 - 5i| + |z + 2 - 5i| < \epsilon$

$|z + 2 - 5i| < \frac{\epsilon}{2}$

$|z - 2 + 4 - 4i - i| < \frac{\epsilon}{2}$

$|(z - 2 - 4i) + (4-i)| < \frac{\epsilon}{2}$

$|z - (2 + 4i)| +|4-i| < \frac{\epsilon}{2}$

$|z - (2 + 4i)| < \frac{\epsilon}{2} - \sqrt{17}$

If I let $\delta = \frac{\epsilon}{2} - \sqrt{17}$, isn't possible that it could be negative?

 

[Bashyboy]

And how might one make this argument more rigorous and convincing, as I am not exactly convinced.

 

[Mark44]

And how might one make this argument more rigorous and convincing, as I am not exactly convinced.

f(z) =5i
Prove that $\lim_{z \to z_0}f(z) = 5i$
Let $\epsilon > 0$ be given.
$|f(z) - f(z_0)| < \epsilon$
$\Rightarrow |5i - 5i| = 0 < \epsilon$
Since $|f(z) - 5i| < \epsilon$ for any choice of z, take $\delta$ equal to whatever you want.

Proofs like this one don't get any simpler than this!

 

[RUber]

$|(5i-z-2) + (z-5i+2)| < \epsilon$

Using the triangle inequality,

$|5i-z-2| + |z-5i-2| < \epsilon$

Be careful here. The triangle inequality states $|a + b| \leq |a| + |b|$
Thus, the right side might be larger in modulus than $\epsilon$.
Making $|a| + |b| < \epsilon$ not necessarily true.