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?

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    $\begingroup$ 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

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.

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  • $\begingroup$ 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


$$\lim_{n \to \infty} f(n)g(n) = \left( \lim_{n \to \infty} f(n) \right) \left(\lim_{n \to \infty} g(n) \right)$$ 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.

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