
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<xa<delta ==> f(x)c>(1/2)bc.
Not really sure how to get started on this one. I'm assuming I need to find a relation between f(x)c and bc but I can't seem to see how to do that.

You do know the definition of "limit", don't you? If $\displaystyle \lim_{x\to a} f(x)= b$ then, for any $\displaystyle \epsilon> 0$, there exist $\displaystyle \delta> 0$ such that if $\displaystyle x a< \delta$ then $\displaystyle f(x) b< \epsilon$. For any c not equal to b, take $\displaystyle \epsilon= b c/2$.
(The fact that this involves complex numbers is irrelevant. Exactly the same proof works if your set is complex numbers or real numbers.)

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?

Let $\displaystyle \varepsilon = \dfrac{{\left {c  b} \right}}{2} > 0$.
From the definition $\displaystyle \left( {\exists \delta>0 } \right)\left[ {\left {x  a} \right < \delta \, \Rightarrow \,\left {b  f(x)} \right < \varepsilon } \right]$
Suppose that $\displaystyle \left( {\exists d} \right)\left[ {\left {d  a} \right < \delta \;\& \,\left {c  f(d)} \right < \frac{{\left {c  b} \right}}
{2}} \right]$.
Note that means $\displaystyle \left {c  b} \right \leqslant \left {c  f(d)} \right + \left {f(d)  b} \right $
There is a contradiction there. What is it?

well...here is what I see so far...
Since bf(x)<epsilon and cf(d)<cb/2=epsilon
then bf(x) + cf(d) should be less then to two times epsilon= cb. But now that I look at this closer I'm not sure that is actually revelant with the statements that you made.

$\displaystyle \left {c  b} \right \leqslant \left {c  f(d)} \right + \left {f(d)  b} \right< \frac{cb}{2}+\frac{cb}{2}=cb$
That is a contradiction. No number is less than itself.

Oh... I see it in now. Alright, that makes a lot of sense. Thanks a lot.