# Limits of Functions

• Oct 16th 2010, 07:29 PM
zebra2147
Limits of Functions
Let f:D-->C(complex numbers).
Prove that if f(x)-->b as x-->a and c=/b, then there exists a delta>0, such that for any x contained in D, 0<|x-a|<delta ==> |f(x)-c|>(1/2)|b-c|.

Not really sure how to get started on this one. I'm assuming I need to find a relation between |f(x)-c| and |b-c| but I can't seem to see how to do that.
• Oct 17th 2010, 07:24 AM
HallsofIvy
You do know the definition of "limit", don't you? If $\lim_{x\to a} f(x)= b$ then, for any $\epsilon> 0$, there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- b|< \epsilon$. For any c not equal to b, take $\epsilon= |b- c|/2$.

(The fact that this involves complex numbers is irrelevant. Exactly the same proof works if your set is complex numbers or real numbers.)
• Oct 17th 2010, 09:27 AM
zebra2147
I did know the definition of limit. The trouble that I was having was with choosing the proper epsilon that makes the proof true. So would you please tell me how you chose the epsilon that you chose?
• Oct 17th 2010, 10:08 AM
Plato
Let $\varepsilon = \dfrac{{\left| {c - b} \right|}}{2} > 0$.
From the definition $\left( {\exists \delta>0 } \right)\left[ {\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {b - f(x)} \right| < \varepsilon } \right]$

Suppose that $\left( {\exists d} \right)\left[ {\left| {d - a} \right| < \delta \;\& \,\left| {c - f(d)} \right| < \frac{{\left| {c - b} \right|}}
{2}} \right]$
.

Note that means $\left| {c - b} \right| \leqslant \left| {c - f(d)} \right| + \left| {f(d) - b} \right|$
There is a contradiction there. What is it?
• Oct 17th 2010, 01:39 PM
zebra2147
well...here is what I see so far...
Since |b-f(x)|<epsilon and |c-f(d)|<|c-b|/2=epsilon
then |b-f(x)| + |c-f(d)| should be less then to two times epsilon= |c-b|. But now that I look at this closer I'm not sure that is actually revelant with the statements that you made.
• Oct 17th 2010, 01:46 PM
Plato
$\left| {c - b} \right| \leqslant \left| {c - f(d)} \right| + \left| {f(d) - b} \right|< \frac{|c-b|}{2}+\frac{|c-b|}{2}=|c-b|$
That is a contradiction. No number is less than itself.
• Oct 17th 2010, 01:53 PM
zebra2147
Oh... I see it in now. Alright, that makes a lot of sense. Thanks a lot.