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The answer is obviously No, with an example: $\frac{1}{n} > 0$ for all n, and $lim \frac{1}{n} = 0$ (I can prove this using delta-epsilon method or just draw a picture).

But I can't wrap my head around this logic. If it is positive for ALL integer n (no matter how large n is), why would the statement not hold at limit, since "limit" just means "n is getting larger and larger" (but still an integer) and infinity is just a symbol, not an actual number?

Can you explain this intuitively for me?

Also, do you have any advices for when I should be careful with this issue: some properties may be true for all elements in a sequence, but maybe false at limit?

(I did this many times in my homework. Whenever I want to claim something about the $\lim a_n$, I always said something like "since it is true for all $a_n$, it must be true for $\lim a_n$" and didn't receive any credits!)

Thanks!

user262959
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    It is helpful to re-read the definition of limit. – Yes Aug 25 '15 at 11:59
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    Another example: every ratio of consecutive Fibonacci numbers is rational, but the limit of this (golden) ratio is not. (Which is handy, since it gives a nice family of rational approximations to the golden ratio.) – Semiclassical Aug 25 '15 at 12:01

3 Answers3

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The intuition for me is that limits are used to define the closure of a set, and so the conditions which persist in the limit are closed conditions.

The example you give with $\frac{1}{n} > 0$ is an open condition. If you replace it with the closed condition $\frac{1}{n} \geq 0$, then the property persists in the limit.

Spencer Leslie
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If the property describes a closed set, then it goes through the limit. This might even BE the definition of closed set.

$\{x \in \Bbb R \mid x > 0\}$ is not a closed set, so you might have problems with limits.

$\{x \in \Bbb R \mid x \ge 0\}$ is a closed set, so if a sequence of positive reals has a limit then it has to be positive or zero.

mercio
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  • Well, I do not think the OP knows some topology of $\mathbb{R}^{n}$ without knowing the concept of limit... – Yes Aug 25 '15 at 12:00
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Limit does rather mean something along the line of "the elements of your sequence (1/n) get closer and closer to the limit, but do not necessarily have to reach the limit". The important thing is that the limit is not necessarily "part of the sequence". Thus, if you can make a statement which holds (only) for all elements of the sequence, it does not necessarily hold for the limit, exccept the limit is part of the sequence.

Consider the following sequence of rational numbers: 3, 31/10, 314/100, 3141/1000, ... which converges to $\pi$. However, albeit all elements of the sequence are rational numbers, the limit $\pi$ is not.

What you actually showed was that the set $(0,\infty)$ (your inequality) is not complete (given the usual metric), and what you constructed was a so called Cauchy sequence (1/n) which does not converge (in $(0,\infty)$), albeit all elements of the sequence are in $(0,\infty)$.

In general, you can not assume any property about the limit of a sequence only because the elements of the sequence have the property. In fact, you cannot even assume that a limit exists at all. This statement does of course not hold if the sequence contains its own limit, e.g. the sequence 1,1,1,1,1... has as 1 as its limit. Besides such exceptions, I would be always careful.

NeitherNor
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