Solving for \lim_{n\to\infty} (-1)^nsin(1/n)

[zacharyh]

Asked to compute:

$$\lim_{n\to\infty} (-1)^nsin(1/n)$$

I've broken this limit down into:

$$\lim_{n\to\infty} (-1)^n * \lim_{n\to\infty}sin(1/n)$$

I've determined $$\lim_{n\to\infty}sin(1/n) = 0$$

Now I have $$\lim_{n\to\infty} (-1)^n * 0$$

This is where I run into trouble...

Attempting to solve for $$\lim_{n\to\infty} (-1)^n$$:

-I've tried plugging in integers and rational numbers for n. It jumps to -1 and 1 with integers, and spits out complex numbers when I plug in rational numbers.
-I've also tried graphing this function on a calculator to no avail.
-I've also plugged it into maple and it spits out: (-1..1).

Is it safe to say $$\lim_{n\to\infty} (-1)^n$$ does not exist?

In which case, I have something that does not exist multiplied by 0, and anything multiplied by 0 equals 0... but I have "nothing" not "anything" ;)

 

[HallsofIvy]

Yes, $\lim_{n\rightarrow \infty} (-1)^n$ does not exist. However,
[tex]- sin(1/n)\le (-1)^nsin(1/n)\le sin(1/n)[/itex]<br /> and sin(1/n) goes to 0 as n goes to infinity. What does that tell you?[/tex]

 

[zacharyh]

Oh I see...sammich' theorem!
How do you produce that inequality?
I'm not sure how to work with sin(1/x)... can it be related to sin(x) somehow?

 

[konthelion]

Oh I see...sammich' theorem!
How do you produce that inequality?
I'm not sure how to work with sin(1/x)... can it be related to sin(x) somehow?

For $$h(x) \leq f(x) \leq g(x)$$

where $$\lim_{n\to\infty} h(x) = L,\lim_{n\to\infty} g(x)=L$$

then

$$\lim_{n\to\infty} f(x)=L$$

Look at HallofIvy's hint. What is L?

 

[zacharyh]

Yes I'm aware that the limit is sandwiched between two 0s and is therefore 0. I guess what I am asking for is a proof of the inequality. How did you decide to pick sin(1/n)? How do I know that it is between those functions?

 

[HallsofIvy]

What? Your question was about (-1)ⁿ sin(1/n). n is either even or odd. (-1)ⁿ is either 1 or -1. I "picked" sin(1/n) because that was the function multiplied by (-1)ⁿ.