Limit of an absolute sequence

[ohreally1234]

1. Homework Statement
Consider the sequence a_n = abs(sin(x))^(1/x)
Find the lim a_n if it exists


2. Homework Equations

None. This is for my calc 2 class.

3. The Attempt at a Solution

We are studying the sandwich theorem, so I thought 0 < M^(1/x) < abs(sin(x))^(1/x) < 1^(1/x).
(Because I assumed that sequences imply x = 1, 2, 3, 4 ..., so sin(x) never equals 0).

Since M^(1/x) and 1 both tend to 1, I reasoned a_n must go to 1.

 

[jhicks]

Do you mean $$a_{n}=|sin(n)|^{\frac{1}{n}}$$? Second, $$\sqrt[x]{1}$$ does not tend to 0 as x becomes large, nor does even $$\sqrt[x]{\frac{1}{2}}$$. That doesn't make using 1 any less valid, you just made an incorrect assumption.

 

[ohreally1234]

sorry i meant that lim goes to 1. (fixed the typo in original post)

 

[jhicks]

You're almost there, but I'm not convinced. Can you prove the existence of an M and that $$\sqrt[x]{M}$$ goes to 1 as x becomes large? I'm not sure how much rigor is required in your class.

 

[ohreally1234]

I reasoned that M exists because the real numbers are dense.
and you can prove M^(1/x) goes to 1 using the definition of the limit.
Are there any holes in my argument?

 

[quadraphonics]

(Because I assumed that sequences imply x = 1, 2, 3, 4 ..., so sin(x) never equals 0).

Well, it won't equal *exactly* 0, but doesn't the density of the reals imply that it gets arbitrarily close? What is $\liminf_{n\to\infty}\left|\sin(n)\right|$?

 

[ohreally1234]

hmm yeah i thought that part of my argument was a bit shady.
can anyone offer some insights?

 

[HallsofIvy]

I reasoned that M exists because the real numbers are dense.

The real numbers are dense in what?

and you can prove M^(1/x) goes to 1 using the definition of the limit.
Are there any holes in my argument?