Homework Help: Simple Limit

1. Jul 30, 2008

zacharyh

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

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" ;)

2. Jul 30, 2008

HallsofIvy

Yes, $\lim_{n\rightarrow \infty} (-1)^n$ does not exist. However,
$$- sin(1/n)\le (-1)^nsin(1/n)\le sin(1/n)[/itex] and sin(1/n) goes to 0 as n goes to infinity. What does that tell you? 3. Jul 30, 2008 zacharyh Oh I see...sammich' theorem! How do you produce that inequality? I'm not sure how to work with sin(1/x)... can it be related to sin(x) somehow? 4. Jul 30, 2008 konthelion For [tex]h(x) \leq f(x) \leq g(x)$$

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?

5. Jul 30, 2008

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?

6. Jul 30, 2008

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.