In Elementary Real Analysis by Bruckner, Bruckner, and Thomson, it is stated on page 187 that right hand limit exists if every decreasing sequence satisfies $f(e_n)/to L, with the other criteria being the usual definition of there limit in ethers of sequences. This is important to me because it determines whether I can or cannot currently prove that a bounded monotonic sequence has a limit, as explored in this post: Prove that if $f$ is bounded and nondecreasing on $(a,b)$ then lim $f(x) $as $x$ approaches $b$ from the left exists.. In my analysis book by Sherbert and Bartle, it is stated on page 112 that all sequences not just monotonic sequences must work. Is the definition standardized? Any help is much appreciated. I have been wrestling with this proof for awhile now.
-
After getting some help on a related question, I suspect the first book watered down the requirement for beginners. – NS248 Mar 18 '14 at 21:58
-
The way you can define one sided limits in terms of sequences usually relies on the how they are defined in terms of $\epsilon - \delta$. For example the left limit is defined just as the usual limit except that, if $x \to x_0^-$ then $x$ is in the nbhd to the left of $x_0$ only. I.e $x\in (x_0-\delta, x_0)$ – user10444 Mar 18 '14 at 22:11
-
Can you give a link in Google books to Bartle's book in which he says any sequence will work? I don't think a decreasing sequence will give you the left limit, if we're considering any sequence. – user10444 Mar 18 '14 at 22:14
-
I was probably too brief. Bartle did include the necessity that $x_n\gt x$ where $x$ is the limit for all $n$. I am still trying to find the right balance on this site between giving too little versus too much information. Anyway, thanks for looking at my question. – NS248 Mar 19 '14 at 00:56
1 Answers
Here are two definitions are right-hand limits: let $f: \mathbb{R} \rightarrow \mathbb{R}$ and let $a \in \mathbb{R}$. Then $ \lim_{x \downarrow a} f(x) = L \quad \textrm{if}$
$$\forall (x_n)_{n=1}^{\infty} \subset (a,\infty) \quad \textrm{such that} \quad \lim_{n \rightarrow \infty} x_n = a \quad \textrm{we have} \quad \lim_{n \rightarrow \infty} f(x_n) = L. \quad (1)$$
$$ \forall (x_n)_{n=1}^{\infty} \subset (a,\infty) \quad \textrm{such that} \quad x_n \downarrow a \quad \textrm{as} \quad n \rightarrow \infty \quad \textrm{we have} \quad \lim_{n \rightarrow \infty} f(x_n) = L. \quad(2)$$
You seem to be worried that these definitions are 'different'. In fact, they are equivalent. Clearly $(1) \Rightarrow (2)$. So let us show $(2) \Rightarrow (1)$:
Take an arbitrary sequence $(x_n)_{n=1}^{\infty} \subset (a,\infty) \quad \textrm{such that} \quad \lim_{n \rightarrow \infty} x_n = a$. Assuming (2) we must show that $\lim_{n \rightarrow \infty} f(x_n) = L.$ To do so we assume the following:
Lemma Suppose $(x_n)_{n=1}^{\infty} \subset \mathbb{R}$ has the following property: $\exists L \in \mathbb{R}$ such that from any subsequence $(x_{n_k})_{k=1}^{\infty}$ we can extract a subsequence $(x_{n_{k_r}})_{r=1}^{\infty}$ with limit $L$ as $r \to \infty$. Then in fact $\lim_{n \to \infty} x_n = L$.
See this page for proof of this Lemma: Sufficient condition for convergence of a real sequence
Now take an arbitrary subsequence $(x_{n_k})$ of $(x_n)$. You should be able to show (by induction) that we may extract a subsequence $(x_{n_{k_r}})$ of $(x_{n_k})$ which approaches $a$ monotonically from above as $r \to \infty$. But now by (2) we have $$\lim_{r \rightarrow \infty} f(x_{n_{k_r}}) = L.$$
We have verified the hypothesis of the Lemma for the sequence $(f(x_n))_{n=1}^{\infty}$ and the required convergence follows.
-
Thanks. I see the equivalence now. Analysis is great, but it's going to take me some time to internalize, clearly. – NS248 Mar 19 '14 at 00:58
-