How to evaluate the limit of a sequence using definition?

Question

To evaluate the limit of a sequence $ \{a_n\}$, we use the tactic of evaluating the expression $ \lim_{n \to \infty}a_n $. Also, at times, when we have a guess of the limit (say $l$) of a sequence, we can try to prove that $l$ is the limit of that sequence, using the definition of limit.

However, I would like to know a method of finding the limit of the sequence using definition. That is, my question is :

Evaluate the limit of the sequence $\left(\frac 1 n\right)$ using the definition of limit.

and not:

Prove that $0$ is the limit of the sequence $\left(\frac 1 n\right)$ using the definition of limit.

Is there any procedure to do this? Other than guessing the limit and then proving it.

Answer 1

Let $\varepsilon>0$ and set $$ n_0=\left\lfloor\frac{1}{\varepsilon}\right\rfloor+1. $$ Clearly $n_0>1/\varepsilon$.

Then for every $n\ge n_0$, $$ \lvert a_n\rvert = \frac{1}{n}\le\frac{1}{n_0}<\varepsilon. $$

Answer 2

There can't be a general technique. Your profile says you do computer science, so maybe you'll like this proof:

Let $L$ be a Turing Machine that, given a sequence, computes the limit to precision $P$ (obviously, we can't always compute the exact value in finite time). We can show this is equivalent to the halting problem. For any machine $M$ and input $I$, define the sequence $a_i$ to be $2P$, if $M$ run on $I$ has halted at the $i$th step, and $0$ otherwise.

Since the machine either halts, and then remains halted, or never halts, the limit of $a_i$ is either $2P$ or $0$. Since $L$ computes the limit to precision $P$, we'll know which one of the two the limit really is. But that will also tell us if $M(I)$ halts. Therefore, no such $L$ can exist.

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On a more useful note: just pass limits through continuous functions, use L'Hopital, and see what happens when you replace $a^b$ with $\exp(b \ln a)$. Those three will probably get you through most problems.