1
$\begingroup$

Where $a$ and $b$ are constants.

I can think of it two different ways. First is that as $n$ goes to infinity, $\frac{b}{n}$ goes to $0$, so that we end up with $\lim_{n \to \infty}\frac{a}{\frac{b}{n}} = \frac{a}{0}$, which is undefined.

The other way is to say that $$\lim_{n \to \infty}\frac{a}{\frac{b}{n}} = \lim_{n \to \infty} a \cdot\frac{n}{b} = a \cdot \infty = \infty$$.

Which one is correct and why?

$\endgroup$
  • 1
    $\begingroup$ I think that the first thought is not correct as the concealed assumption is that $lim_{n\to \infty}\frac{a}{\frac{b}{n}}=\frac{lim_{n\to \infty}a}{{lim_{n\to \infty}} \frac{b}{n}}$ $\endgroup$ – gbox Jul 18 '16 at 6:49
  • $\begingroup$ Your second approach is correct. Note that if $\frac{a}{b} < 0$ the limit goes to $- \infty$, while if $\frac{a}{b} = 0$ then it is $0$. On the other hand, if $b = 0$, then the expression is undefined, since it is undefined for every $n$. $\endgroup$ – Shagnik Jul 18 '16 at 6:50
  • $\begingroup$ When trying to answer such a question, the first step is not to think in terms of limit "formulas," but instead think about what's happening, for various concrete $a$ and $b$. $\endgroup$ – André Nicolas Jul 18 '16 at 6:54
  • $\begingroup$ @AndréNicolas, yes that's what I was trying to do and why I thought it was undefined, but apparently that's wrong. I thought if you have some constant divided by another constant which is itself divided by something growing to infinity, then that fraction $\frac{b}{n}$ is going to go to $0$ and therefore the whole thing will be undefined. $\endgroup$ – jeremy radcliff Jul 18 '16 at 6:56
  • $\begingroup$ The issue then was perhaps of thinking of $n$ somehow as "infinite" instead of thinking of it as very large. $\endgroup$ – André Nicolas Jul 18 '16 at 6:59
2
$\begingroup$

However it is written at the outset we are given the sequence $$x_n:={a\,n\over b}\qquad(n\geq1)\ ,$$ with the tacit assumption that $b\ne0$. If $a=0$ then $x_n=0$ for all $n$, hence $\lim_{n\to\infty}x_n=0$. If $a\ne0$ then we all know that the $x_n$ converge to $\infty$ if $ab>0$, and to $-\infty$, if $ab<0$. This means that the sequence is divergent in ${\mathbb R}$. Nevertheless we are entitled to write $$\lim_{n\to\infty}x_n=\infty\quad(ab>0),\qquad \lim_{n\to\infty}x_n=-\infty\quad(ab<0)\ ,$$ meaning that we accept $\pm\infty$ as limiting values, and have verified the corresponding convergence conditions.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Is it ok to think of sequences with real number indices, or are we thinking of the sequence $x_n$ with integer values for $n$ assuming that the general behavior will hold for real number values of $n$? I always thought of sequence indices as integer values. $\endgroup$ – jeremy radcliff Jul 18 '16 at 8:17
  • $\begingroup$ @jeremyradcliff: The domain of an (infinite) sequence is ${\mathbb N}$, beginning with $0$ or $1$. When you write $x_n$ everyone assumes automatically that $n$ runs through a set of natural numbers. If we prove something about such a sequence we do not claim that the statement also holds when the $n$ in the expression defining $x_n$ is interpreted as a real number. E.g., $\lim_{n\to\infty}\sin(n\pi)=0$, but $\lim_{x\to\infty}\sin(\pi x)$ does not exist. $\endgroup$ – Christian Blatter Jul 18 '16 at 10:02
  • $\begingroup$ Thank you for the detailed explanation, this is very helpful. $\endgroup$ – jeremy radcliff Jul 18 '16 at 12:46
0
$\begingroup$

Well,

$$\lim_{n \to \infty} f(n)g(n) = \left( \lim_{n \to \infty} f(n) \right) \left(\lim_{n \to \infty} g(n) \right)$$ only holds if both $\lim_{n \to \infty} f(n)$ and $\lim_{n \to \infty} g(n)$ exist.

In this case, $f(n)=a$ and $g(n) = \frac{n}{b}$. It is clear then that $\lim_{n \to \infty} g(n) $ does not exist.

| cite | improve this answer | |
$\endgroup$

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.