#### 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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