Proving The Continuous Theorem for Sequences

1. May 10, 2007

Lucretius

1. The problem statement, all variables and given/known data
4.8 Show the following continuous theorem for sequences: if $$a_n \rightarrow L$$ and f is a real valued function continuous at L, then $$bn = f(a_n) \rightarrow f(L)$$.

2. Relevant equations

No real relevant equations here. Just good old proof I'm thinking.

3. The attempt at a solution

Well, I stared at this for an hour today. I was able to complete the rest of the assignment but this one has me stumped. I realize that $$\displaystyle\lim_{n\rightarrow\infty}a_n=L$$ and that for a real-valued function to be continuous at L that $$\displaystyle\lim_{x\rightarrow x_0}f(x)=f(x_0)=L.$$ I don't know what to do from here though. How do I get f(L) from f(x0)=L, and then get f(a_n) from just plain old a_n. This thing makes intuitive sense to me; it's blatantly obvious it's right - proving it has ... well.. proven to be really hard!

2. May 10, 2007

NateTG

When something seems like it should be obvious, you might want to write out some definitions. Epsilons, deltas, and all that.

3. May 10, 2007

Lucretius

Well that's the thing; I thought I could just approach it nicely, but the concepts seem to be two different things; it's as if I want to prove a finite limit at infinity. There is a proof I thought was similar to it, in the notes he gave us. However, I fleshed out a few epsilon delta definitions; drew a nice picture so I could see what was going on. My main problem is I have no clue how to get to f(a_n) from a_n, and from f(x0)=L to f(L).

4. May 11, 2007

NateTG

It looks like you might be getting confused by using two different $L$'s. Typically, the $a_n$ and $b_n$ will not converge to the same value.

Really, this is straightforward stuff... given some $\epsilon$ greater than $0$ can you show that there is some $N$ so that $n>N \Rightarrow |f(a_n)-f(L)| < \epsilon$?

5. May 11, 2007

Lucretius

Hmm, I must not be seeing it...

I get that it's going to follow the basic form of a proof. For all epsilon > 0, there exists an N such that, for all n>N, then |f(a_n)-f(L)|<epsilon. The N is more than likely going to be some integer part of some number plus one. I get how all that works. The problem is... I'm used to an |f(x)-L|, where i know what the L and f(x) are. I have no clue as to what f(L) or f(a_n) are, and I don't see how I can understand what they are from the given information. Sorry; I hate missing obvious stuff =(

6. May 11, 2007

NateTG

It's a little abstract. Do you know if there's a $\delta$ so that:
$$|a_n-L|<\delta \Rightarrow |f(a_n)-f(L)| < \epsilon$$

7. May 11, 2007

Lucretius

I'm following you and it looks like I'll be proving a limit at a point, but I'm still stuck with how we can relate a_n to f(a_n), and L to f(L). If I can somehow extract the a_n and the L from the f, I could do it...again, thanks for the help so far and sorry I'm still not quite getting what to do...

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