What's the Issue with the Second Approach to Solving the Limit of Sinx /x?

[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?

 

[pasmith]

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?

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

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