- #1
ChrisVer
Gold Member
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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?
[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?
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