Limit of function ( sandwich method)

[Дьявол]

limit of function ("sandwich" method)

Homework Statement

Using the "sandwich" method prove that $$\lim_{n\rightarrow \propto }(\frac{sin(n)}{n})=0$$

Homework Equations

$$x_n \leq y_n \leq z_n$$

$$\lim_{n\rightarrow \propto }(x_n) \leq \lim_{n\rightarrow \propto }(y_n) \leq \lim_{n\rightarrow \propto }(z_n)$$

The Attempt at a Solution

I am honestly little bit confused at this point.

If the answer is:

$$\frac{-1}{n} \leq \frac{sin(n)}{n} \leq \frac{1}{n}$$

then my question is if $n=-\frac{\pi}{4}$ then $$\frac{-1}{-0.785}$$ will be not less or equal to $$\frac{\sqrt{2}}{2*(-0.785)}$$, where -0.785=$-\frac{\pi}{4}$, where $\pi \approx 3.14$.

Thanks in advance.

 

[marcusl]

Are you sure that n is a real number? Usually n denotes a positive integer in this type of problem.

 

[HallsofIvy]

$-1/n\le sin(n)/n\le 1/n$ for n positive. Obviously, if n is negative, just [itex]-1/n\le 1/n[itex]is not true! Your use of $x\rightarrow \propto$ is a little confusing. Did you mean $\infty$? Even if you do not interpret n as necessarily being positive, if n is "going to $\infty$" eventually, for some finite N, if n> N, n will be positive. And you can always drop any finite number of terms in an infinite sequence without changing the limit.[/itex][/itex]

 

[Дьявол]

Thanks for the posts. I see now, it was my mistake if a_n=sin(n)/n, a_n is progression where n are positive integer numbers. So if:
$$-1 \leq sin(n) \leq 1$$
then divided by n, I'll get:
$$-1/n \leq sin(n)/n \leq 1/n$$
Sorry for the symbol, I misspelled it, since I don't cover LaTeX too good at this moment.
Thanks for the help.

 

[HallsofIvy]

For future reference, in LaTex, $\infty$ is "\infty". $\propto$ is "\propto", i.e. "proportional to".