Sinx /x limit

1. Feb 15, 2014

ChrisVer

Hello, I am just having a small confusion in extraction of the limit:
$lim_{x→∞} \frac{sinx}{x}$

One way to do it is by the sandwich rule:
$- \frac{1}{x} ≤ \frac{sinx}{x}≤ \frac{1}{x}$
from where you get that by taking the limit on both sides:
$lim_{x→∞} \frac{sinx}{x}=0$

Now on the other hand I'd like to write the sin through one of its definitions:
$sinx=∑_{n=0} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}$
and deal it as a polynomial.... In that case, by taking the limit of x approaching infinity, I will get infinity instead, since every $x^{k}, k>1$ will cancel out the denominator...
In fact it will be like:
$lim_{x→∞} \frac{1}{x} ∑_{n=0} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}$
$lim_{x→∞} \frac{x-x^{3}/6 + ...}{x}$
$lim_{x→∞} (1-x^{2}/6 + ...)$
which in fact is undefined...

What is the problem of the 2nd approach?

Last edited: Feb 15, 2014
2. Feb 15, 2014

pasmith

The problem with the second approach is that if $x$ is sufficiently large then
$$\frac{x^n}{n!} < \frac{x^{n+2}}{(n+2)!}$$
so you can't ignore the tail.

EDIT: more precisely,
$$\lim_{x \to \infty} \lim_{N \to \infty} \sum_{n=0}^N \frac{x^{2n}}{(2n+1)!} = 0$$
but
$$\lim_{N \to \infty} \lim_{x \to \infty} \sum_{n=0}^N \frac{x^{2n}}{(2n+1)!}$$
does not exist.

Last edited: Feb 15, 2014