How to evaluate the limit of a sequence using definition?

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.

• There isn't any procedure that works all the time, but L'hopital's rule is a very useful technique to know. – Henry Swanson Sep 25 '14 at 7:18
• Why has the question been downvoted? – Parth Thakkar Sep 25 '14 at 7:40

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

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

• I think the OP's question was "how do I know what the limit is", and the $\frac{1}{n}$ was just an example. – Henry Swanson Sep 25 '14 at 7:19

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.

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.

• I was asking this just out of curiosity. I do know how to solve limit problems using the regular techniques. – Parth Thakkar Sep 25 '14 at 7:44
• And going through my profile before answering the question was cool! – Parth Thakkar Sep 25 '14 at 7:45