Sinx /x limit

ChrisVer

Gold Member
3,330
434
Hello, I am just having a small confusion in extraction of the limit:
[itex] lim_{x→∞} \frac{sinx}{x}[/itex]

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

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


What is the problem of the 2nd approach?
 
Last edited:

pasmith

Homework Helper
1,631
351
Hello, I am just having a small confusion in extraction of the limit:
[itex] lim_{x→∞} \frac{sinx}{x}[/itex]

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

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


What is the problem of the 2nd approach?
The problem with the second approach is that if [itex]x[/itex] is sufficiently large then
[tex]
\frac{x^n}{n!} < \frac{x^{n+2}}{(n+2)!}
[/tex]
so you can't ignore the tail.

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

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