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Can anyone explain this theorem?

I have been read on limits and come across this link between the limit of a function and the convergence of sequence but I cant understand what it is saying fully.

I think this is maybe to do with my english knowledge maybe so I find it to hard to read.

Could anyone explain in basic what it is saying please.

Thanks

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Reply 1
Avatar for poorform
poorform
OP
10
I think it is saying that the limit exists if and only if all sequences that converge to c have the same limit at f(p_n) and p_n can never be c but can get infinitely closer to c.
Original post by poorform
I think it is saying that the limit exists if and only if all sequences that converge to c have the same limit at f(p_n) and p_n can never be c but can get infinitely closer to c.

Yes. "Continuity iff the function preserves the convergence of sequences". It's the reason you are allowed to say things like lim⁡x→0exp⁡(x)=exp⁡(lim⁡x→0x)=exp⁡(0)=1\displaystyle \lim_{x \to 0} \exp(x) = \exp(\lim_{x \to 0} x) = \exp(0) = 1: because exp⁡\exp is continuous, so the first equality sign is justified.

This theorem is what lets us interchange limits with continuous functions.
Original post by poorform
Could anyone explain in basic what it is saying please.



Suppose you have a function f(x)f(x) and you want to know if lim⁡x→cf(x)\displaystyle \lim_{x \rightarrow c} f(x) exists i.e. that lim⁡x→cf(x)=L,L∈R\displaystyle \lim_{x \rightarrow c} f(x) = L, L \in \mathbb{R}.

Then this is true if for *every* sequence xnx_n such that lim⁡n→∞xn=c\displaystyle \lim_{n \rightarrow \infty} x_n = c, we have lim⁡n→∞f(xn)=L\displaystyle \lim_{n \rightarrow \infty} f(x_n) = L

Conversely if lim⁡x→cf(x)=L,L∈R\displaystyle \lim_{x \rightarrow c} f(x) = L, L \in \mathbb{R}, then it is true that lim⁡n→∞f(xn)=L\displaystyle \lim_{n \rightarrow \infty} f(x_n) = L for *every* sequence xnx_n such that lim⁡n→∞xn=c\displaystyle \lim_{n \rightarrow \infty} x_n = c

This theorem is also true for the limit at infinity i.e. if we write c=∞c=\infty in the statements above.

So consider f(x)=sin⁡πxf(x) = \sin \pi x. Then we can see that the limit at infinity of the function does *not* exist by considering what happens to the function at the points defined by xn=n,yn=n+12x_n = n, y_n=n+\frac{1}{2} where both xn,yn→∞x_n,y_n \rightarrow \infty.

I suspect that it is easier to use this theorem to prove that the limit at infinity doesn't exist (as my example shows) that to show that it does exist.

On the other hand, once you know that, say, lim⁡x→∞f(x)\displaystyle \lim_{x \rightarrow \infty} f(x) exists, then you can use it to find the limit of a sequence e.g.

lim⁡x→∞ln⁡(x+1)x=0\displaystyle \lim_{x \rightarrow \infty} \frac{\ln(x+1)}{x} = 0 by L'Hopital's rule, so:

lim⁡n→∞ln⁡(n+1)n=0\displaystyle \lim_{n \rightarrow \infty} \frac{\ln(n+1)}{n} = 0 by setting xn=n→∞x_n=n \rightarrow \infty

lim⁡n→∞ln⁡(ln⁡(n)+1)ln⁡(n)=0\displaystyle \lim_{n \rightarrow \infty} \frac{\ln(\ln(n)+1)}{\ln(n)} = 0 by setting xn=ln⁡(n)→∞x_n= \ln(n) \rightarrow \infty
Original post by atsruser
I suspect that it is easier to use this theorem to prove that the limit at infinity doesn't exist (as my example shows) that to show that it does exist.


There are some easy examples: the constant function f(x)=0f(x) = 0 has the limit 0 at infinity, because for every xn→∞x_n \to \infty we have (f(xn))n≥1=(0)n≥1→0(f(x_n))_{n \geq 1} = (0)_{n \geq 1} \to 0.
Reply 5
Avatar for a10
a10
19
That mathematics looks like chinese to me :lol:
Original post by a10
That mathematics looks like chinese to me :lol:


Most disturbing.
Original post by Smaug123
There are some easy examples: the constant function f(x)=0f(x) = 0 has the limit 0 at infinity, because for every xn→∞x_n \to \infty we have (f(xn))n≥1=(0)n≥1→0(f(x_n))_{n \geq 1} = (0)_{n \geq 1} \to 0.


Are there any non-easy examples? That's just a little too easy for my liking.
Hhahhahahha.

What language is this????
Original post by atsruser
Are there any non-easy examples? That's just a little too easy for my liking.


lim⁡x→∞f(x)\displaystyle \lim_{x \to \infty} f(x) exists, for any ff increasing and bounded above. Indeed all sequences xn→∞x_n \to \infty have f(xn)→xf(x_n) \to x, some xx depending on (xn)n(x_n)_n.

Take the sup over all those xx. The sup exists, because ff is bounded above so none of the limits can be bigger than that bound. Call that sup LL.

Claim: any (xn)→∞(x_n) \to \infty has f(xn)→Lf(x_n) \to L.
Proof: we just need to show that the limit is not less than LL. Suppose it were, calling it xx, and take a collection of sequences (ynm)n≥1(y^m_n)_{n \geq 1} for each m∈Nm \in \mathbb{N}, such that ynm→ymy^m_n \to y^m as n→∞n \to \infty with ym>L−1my^m > L-\frac{1}{m}. (We can do that: LL is a sup.)
Then fix the least mm such that ynm→ym>xy^m_n \to y^m > x. Henceforth abbreviate the sequence ynm→ymy^m_n \to y^m by yn→yy_n \to y.

Summary: xn,yn→∞x_n, y_n \to \infty; f(xn)→x<yf(x_n) \to x < y with f(yn)→yf(y_n) \to y.

But ff is increasing, so we have f(k)≤x<yf(k) \leq x < y for arbitrarily large kk, whence all kk have f(k)≤x<yf(k) \leq x < y. This contradicts that f(yn)→yf(y_n) \to y.
(The f(k)≤xf(k) \leq x is true because ff is increasing, so we can't be tending to the limit from above.)

Therefore all sequences xnx_n have f(xn)f(x_n) tending to the same limit.

Is that nontrivial enough? I know it's kind of obvious, but I don't actually know if there's an easier proof of the statement.

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