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 -1/n\le 1/nis 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 postive. And you can always drop any finite number of terms in an infinite sequence without changing the limit.

 

[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".