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If I'm told to prove that $\lim \limits_{n \to \infty}{\frac{1}{n^2}} = 0$, I would do the following:

$|\frac{1}{n^2} - 0| = \frac{1}{n^2} < \epsilon \implies N(\epsilon) = \frac{1}{\sqrt{\epsilon}} < n$

Every $\epsilon > 0$ is defined, so $\frac{1}{n^2} \to \infty$.

However, what if I'm not told $L = 0$? How would I find that $L = 0$, without looking at graphs or testing increasingly larger $n$?

I used $a_n = \frac{1}{n^2}$ as a simple example. I'd prefer some method that works for other sequences, too.

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    There's no mechanical method to find the limit, you need to let your intuition flow. For this, many people find useful to graph stuff and test increasingly larger values of $n$. – 2017-02-08
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    @GitGud -- I like using my intuition, but I end up not "showing my work" when I do so. For example, I'd imagine that a teacher/professor would not be pleased if they saw merely "$L = 0$" as the solution to "Find $L$ if $L = \lim\limits_{n \to \infty}{\frac{1}{n^2}}$". – 2017-02-08
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    For this particular example, they'd expect a $\varepsilon$-$N$ proof. For more complicated sequences, the grader would probably be satisfied would a proof that uses basic limit arithmetic and known limits. Basically you need to interpret what is really being asked. – 2017-02-08
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    @SirJony There's a method that could be helpful in some cases: first prove the limit exists, say by showing the sequence is Cauchy. Then choose wisely a nice (infinite, of course) subsequence whose limit is more or less easy to evaluate (or which you already know), then you'll already know your whole sequence's limit. – 2017-02-08
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    You use strict mathematics to check whether a candidate limit is indeed a limit. It's entirely fine to use intuition, guessing and hand-wavey mathematics to find candidates. – 2017-02-08
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    @Arthur -- But what if there was a sequence that is too hard to use intuition; what would someone do then? – 2017-02-08
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    Well, then you have mathematical tools, both strict and hand-waving. Calculation, graphing, testing, guessing. There is no one way to solve'em all. Each new sequence might need its own approach. – 2017-02-08
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    @Arthur -- Alright, but what is the strict tool you mention? $\epsilon$-$N$ is more of a verification than a solution, so I suppose you mean something else. – 2017-02-08
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    Different term manipulations, for instance. Look up the Basel problem for an example. Again, this is usually tricks that work only once, at least without tweaking. – 2017-02-08

2 Answers 2

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Finding/evaluating a limit requires the use of limit theorems like

  • algebra of limits
  • Squeeze theorem
  • Cesaro Stolz theorem
  • standard limits like $1/n \to 0$ as $n\to\infty$ which are proved using definition of limits.

Apart from these techniques sometimes we need to use theorems on existence of limits like a monotone and bounded sequence tends to a limit and then use the this fact to find the limit. Such solutions are often tricky. I show one example.

Let $a_n=x^{n}$ where $x$ is real. I deal with the case when $0

Your question about sequence $1/n^{2}$ can be solved by using product rule and the standard limit $1/n\to 0$. In fact we can generalize this and use the result $1/n^{a}\to 0$ for $a>0$ as a standard limit. Sometimes a combination of the techniques mentioned above may be required.

The above techniques exclusively deal with theorems related to sequences. Sometimes we also need results which deal with limits of functions of a real variable. Two such results which are popular are as follows:

  • If $a_n\to L$ and a function $f$ is continuous at $L$ then $b_n=f(a_n) \to f(L) $.
  • If $f(x) \to L$ as $x\to\infty$ then $f(n) \to L$ and if $g(x) \to L$ as $x\to 0$ then $g(1/n)\to L$.

Thus for example $n\sin(1/n)\to 1$ as $n\to\infty$.

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Prove that $1/n^2$ cannot converge to $L$ when $L\ne 0.$ As follows:

(1). If $L<0$ and $0 <\epsilon <|L|/2$ then $|L-1/n^2|>\epsilon$ for ALL $n$.

(2). If $L>0$ there exists $n_0\in \mathbb N$ with $n_0>\sqrt {2/L}.$ Then with $\epsilon = 2/L$ we have : $\quad |L-1/n^2|>\epsilon$ for all $n\geq n_0$.

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    In your solution, it seems as if you had the "bias" (or were very lucky) that you knew that $L=0$. – 2017-02-08
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    There is no general recipe for finding the limit (or even proving that it exists) for ALL formulas. There are some general results that are useful. E.g : If $f(n)\to L$ then $f(n)^2\to L^2.$ – 2017-02-09