Trigonometric limits tending to infinity

[Spider84]

Hello.
Excuse me for my bad mathematic english.
I dont have any exact problem - I just want to understand the algorith of solving the limits of functions having trigonometroc functions tending to infinity.
I know that limits of simple trig functions tending to infinity is not exists.

 

[JeffM]

Hello.
Excuse me for my bad mathematic english.
I dont have any exact problem - I just want to understand the algorith of solving the limits of functions having trigonometroc functions tending to infinity.
I know that limits of simple trig functions tending to infinity is not exists.

I do not know that a special algorithm for finding the limits of trigonometric functions exists.

\(\displaystyle x \ne 0 \implies - \dfrac{1}{x} \le \dfrac{sin(x)}{x} \le \dfrac{1}{x} \implies\)

\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\left(- \dfrac{1}{x}\right) \le \lim_{x \rightarrow \infty}\dfrac{sin(x)}{x} \le \lim_{x \rightarrow \infty}\dfrac{1}{x} \implies\)

\(\displaystyle \displaystyle 0 \le\lim_{x \rightarrow \infty}\dfrac{sin(x)}{x} \le 0 \implies\)

\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\dfrac{sin(x)}{x} = 0.\)

This is a standard procedure, not unique to trigonometric functions.

 

[Spider84]

I do not know that a special algorithm for finding the limits of trigonometric functions exists.

\(\displaystyle x \ne 0 \implies - \dfrac{1}{x} \le \dfrac{sin(x)}{x} \le \dfrac{1}{x} \implies\)

\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\left(- \dfrac{1}{x}\right) \le \lim_{x \rightarrow \infty}\dfrac{sin(x)}{x} \le \lim_{x \rightarrow \infty}\dfrac{1}{x} \implies\)

\(\displaystyle \displaystyle 0 \le\lim_{x \rightarrow \infty}\dfrac{sin(x)}{x} \le 0 \implies\)

\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\dfrac{sin(x)}{x} = 0.\)

This is a standard procedure, not unique to trigonometric functions.

Thanks.
Can I find somewhere explanation of this procedure?

 

[pka]

Can I find somewhere explanation of this procedure?

If \(\displaystyle f(x)\) is a bounded function then \(\displaystyle \exists B>0\) such that \(\displaystyle \forall x,~|f(x)|\le B\).

That means that \(\displaystyle {\lim _{x \to \infty }}\frac{{f(x)}}{x} = 0\) because \(\displaystyle \frac{{ - B}}{x} \leqslant \frac{{f(x)}}{x} \leqslant \frac{B}{x}\).

 

[JeffM]

Thanks.
Can I find somewhere explanation of this procedure?

You may have misunderstood me. I was not saying that the particular procedure used in my example was the only procedure available for trigonometric functions; I was showing that procedures that work on all sorts of functions will work for trigonometric functions. As for that particular procedure, pka has shown you its general form, and the basic intuitions behind the formalization are very simple.

\(\displaystyle First\ intuition:\ a \le b \le a \implies b = a.\) If the number a is not less than 3 and is not more than 3, then a must be 3.

\(\displaystyle Second\ intuition: 0 < |c| << |x| \implies \dfrac{c}{x} \approx 0.\)

Example

\(\displaystyle c = 1,\ 10 \le |x| \implies -0.1 \le \dfrac{c}{x} \le 0.1.\)

\(\displaystyle c = 1,\ 100 \le |x| \implies -0.01 \le \dfrac{c}{x} \le 0.01.\)

\(\displaystyle c = 1,\ 1000 \le |x| \implies -0.001 \le \dfrac{c}{x} \le 0.001.\)

The function (c / x) is approaching zero as x gets very large relative to c. The rigorous formalization of this intuition is what is meant by the limit as x approaches infinity.