Suppose we have a sequence $x_n = (\frac{1}{n})$ and we want to find $$\lim_{n \to \infty} x_n = \ ?$$ by the definition of limit.
Clearly, the limit is $0$ but if we were not able to determine this by intuition (as is the case with more complex sequences), is there a way to find that the limit of this sequence is $0$ by using the definition of a limit? Or is the only option to make an educated guess and attempt to prove it using the definition?
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$\begingroup$You could say that any attempt to prove convergence fails if one assumes any other limit than $0$. But in a way one can hardly determine a limit in general without resorting to educated guessing - thoigh this concept can hardly be formalized. Let's try a few examples and you decide if one can determine or has to guess the limit:
Consider the Fibonacci sequence $F_0=0$, $F_1=1$, $F_{n+1}=F_n+F_{n-1}$. What about $\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\frac{1+\sqrt 5}2$?
What about $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}6$?
To extend the previous example, whyt do we know about $\sum_{n=1}^\infty\frac1{n^3}$ that's really better than computing a partial sum with more or less summands?
In a similar direction, limits such as $\lim_{n\to\infty}(\ln n-\sum_{k=1}^n\frac1k)$ exist, but they are actually used to define new constants as otherwise there's no way known to "obtain" the value.
Good question. To answer it, I think it will help to notice that the basic definition here is "convergence," rather than "limit":
(1) the sequence $\{x_n : n\in \mathbb{N}\}$ converges to $L$ if ...
where the "..." is a particular property of the sequence $\{x_n: n\in \mathbb{N}\}$ and the number $L$.
One can then prove a given sequence converges to at most one value. Then one defines "convergent sequence," and finally one defines the limit of a convergent sequence: "If $\{x_n: n\in \mathbb{N}\}$ is a convergent sequence then its limit is the unique value of $L$ satisfying (1)."
Only after establishing all of this can we go back and rewrite the definition (1) to say
(2) the limit of the sequence $\{x_n : n\in \mathbb{N}\}$ is $L$ if ...
However, note that (2) cannot serve as the basic definition in place of (1) because, even though at first glance it seems to be phrased as a definition of "limit," it is not clear a priori that it results in a well-defined notion.
Now my point is that to verify some fact about limits from the definitions above, the basic definition that needs to be verified is that of convergence. Since convergence applies to pairs of objects (the sequence $\{x_n : n\in \mathbb{N}\}$ and the number $L$,) if you want to verify it, then you have to supply both of these objects. Presumably the sequence $\{x_n : n\in \mathbb{N}\}$ comes from the problem at hand. Depending on the situation, the limit $L$ might come from your intuition, from an educated guess, or from the conclusion of some theorem. But you can't expect it to come out of the definition of convergence. Rather, it would make more sense to think of it as one of the inputs to the definition.
Note that going from definition (1) to definition (2) doesn't change this; all we have done is to reword things a bit, so we still can't expect this to give us a useful mechanical procedure that takes a description of a sequence as input and gives a simple description of its limit (if it exists) as output. (Of course we could use the "trial and error" procedure of checking whether various values of $L$ satisfy the definition, which, although mechanical, is hardly useful.)
I don't claim to have proved that there is no such mechanical procedure; the notion that finding limits is hard comes rather from experience. But I hope this answer gives some explanation of why we shouldn't expect such a procedure to arise from the definitions.
$\endgroup$ $\begingroup$Here is my go at making it a bit more rigorous:
First note : 1/1 > 1/2,
In general, given 1/n, note (where n is a positive integer):
n + 1 > n
1/(n+1) < 1/n
By induction, we now that every term in this sequence is of smaller magnitude than the previous term.
However, we have not proven that the limit is zero, for there are many sequences, e.g. 2^(1/n), for which the above induction would hold but for which the sequence does not approach 0.
In this case, the limit of 1/n can be found by analyzing the limit of the sequence "n". This is because, if a sequence diverges when take at infinity, we also get to know something about it's reciprocal, namely it must be converging to a very specific value: 0.
To further justify, note a sequence defined by the converging series K. Clearly, the addition of each consecutive term to the series increases the magnitude of the sum by a steadily decreasing bit, however the rate of decrease is set so that a limit exists (consider the series for 2^(1/x) and apply the laws of geometric series to find the limit) . Since this limit takes the form of a number which we can denote by m, its reciprocal must exist (this is based on the axiom of inverse elements). Therefore, the reciprocal 1/m has a definite value (ascertainable or not).
However, in the case of a diverging sequence, i.e. like the natural numbers (the divergence of which relies on the fact that the natural number sequence can be defined by an arithmetic series, so that the difference between each consecutive term, by virtue of definition is constant), there is no upper limit to the highest element n, as in this case, the number n + 1, which can be demonstrated to be a member of this sequence by recursion, would be greater than n.
So, by the same step with reciprocals, no 1/n (where n is an element of the natural numbers) term will be the smallest member of its sequence. Yet it is clear that any magnitude bound, e.g. finding a 1/k smaller than 5^(-43544), can be satisfied by considering something like 1/(1+5^43544). Hence, no bound of magnitude will be the minimal bound, no matter of large the denominator; so finally, the magnitude of this sequence, as it does not converge to a non-0 number (otherwise there would be a bound on magnitude), must be going to well nothing. The idea is that the magnitudes get steadily smaller and smaller, and well, the smallest magnitude that would be approached if we kept doing this for an infinitely long time, would be "nothingness", 0 itself.
Alternatively, we can define the limit at infinity as the magnitude which the sequence will approach, but never reach, under finite conditions. So since any arbitrary positive bound (let's say less than 1) can be satisfied in with finitely many recursions, the ultimate bound which cannot be finitely satisfied must be 0 itself. Hence, the limit is 0.
I think the real point of any "justification" would be to demonstrate that 1/n indeed does not converge to a non-zero positive real number.
Have a nice day! :)
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