Does Continuity Guarantee Half the Function Value in Convergent Sequences?

[sillyus sodus]

Homework Statement

Given a convergent sequence:
a_n \rightarrow a
and a continuous function:
f:\mathbb{R}\rightarrow\mathbb{R}
show that there exists an N\in\mathbb{N} such that \forall n>N:
f(a_n)\geq\frac{f(a)}{2}

Homework Equations

Usual definitions for limit of a sequence and continuous function.

The Attempt at a Solution

I've tried playing around with it but I really don't understand what to do, I know its an easy question and I feel pretty stupid but I'm stuck. Can anyone just get me started on this one?
So far I've rearranged it so it looks like:
f(a)-f(a_n)\leq f(a_n)
Which sort of looks like a limit without the abs values...
I figure that the function of a sequence is a sequence again, btu what am I trying to say here? That it is a divergent sequence?

 

[gammamcc]

False. Let f(x) = -1 (constant function). Let a_n =0 for all n. Here, f(a_n)=f(0)=-1 <-1/2 for all n, yet a_n:n=1,2,... converges in the domain if f.
[Is there another condition?]

 

[sillyus sodus]

Sorry, sorry, yes there is another condition:
f(a)&gt;0

 

[gammamcc]

To not give the answer away, I would just suggest that the problem is now a matter of
fitting into the \epsilon - \delta definition of continuous functions. Note that f(a)/2 < f(a).

 

[HallsofIvy]

Take \epsilon&gt; a/2 in the definition of "continuous function".

 

[gammamcc]

whaaaa.? So much for the 'for all epsilon >0' part. Whaaa? Well, not my problem anyway...

 

[HallsofIvy]

Yes, "For all epsilon> 0", something is true so it is true for any specific value of epsilon. Here we are given that f is continuous so we are free to choose any value of epsilon we like.

 

[sillyus sodus]

I have an answer now, basically because a_n\rightarrow a and f is continuous then f(a_n)\rightarrow f(a), so there exists n bigger than N such that |f(a_n)-f(a)|&lt;\epsilon so then just choose \epsilon = f(a)/2 do some rearranging and the inequality pops out.

In other words, f(a_n) is a convergent sequence.