0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ There isn't any procedure that works all the time, but L'hopital's rule is a very useful technique to know. $\endgroup$ – Henry Swanson Sep 25 '14 at 7:18
  • $\begingroup$ Why has the question been downvoted? $\endgroup$ – Parth Thakkar Sep 25 '14 at 7:40
0
$\begingroup$

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

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

$\endgroup$
1
  • 2
    $\begingroup$ I think the OP's question was "how do I know what the limit is", and the $\frac{1}{n}$ was just an example. $\endgroup$ – Henry Swanson Sep 25 '14 at 7:19
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I was asking this just out of curiosity. I do know how to solve limit problems using the regular techniques. $\endgroup$ – Parth Thakkar Sep 25 '14 at 7:44
  • $\begingroup$ And going through my profile before answering the question was cool! $\endgroup$ – Parth Thakkar Sep 25 '14 at 7:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.