## Proving The Continuous Theorem for Sequences

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!
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 Recognitions: Homework Help Science Advisor When something seems like it should be obvious, you might want to write out some definitions. Epsilons, deltas, and all that.
 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).

Recognitions:
Homework Help

## Proving The Continuous Theorem for Sequences

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 $n>N \Rightarrow |f(a_n)-f(L)| < \epsilon$?
 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)|
 Recognitions: Homework Help Science Advisor It's a little abstract. Do you know if there's a $\delta$ so that: $$|a_n-L|<\delta \Rightarrow |f(a_n)-f(L)| < \epsilon$$
 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...