(x_n)->0 and lim(x_n)sin(1/(x_n))=0 help?

[Unassuming]

(x_n)-->0 and lim(x_n)sin(1/(x_n))=0 ...help?

Homework Statement

Let x_n be a sequence in R with x_n \rightarrow 0 and x_n \neq 0 for all n. Prove that lim (x_n) sin \frac{1}{x_n} = 0.

The Attempt at a Solution

I think I might have a solution if I say that if x_n \rightarrow 0, then \frac{1}{x_n} \rightarrow \infty.

Then sin \frac{1}{x_n} \times (x_n) \rightarrow 0. This might or might not be a good method.

Anyway, ... I am required to use the squeeze thrm to prove this. Could somebody give me a hint on which two outer limits to use?

 

[e(ho0n3]

Are you sure that x_n converges to 0? Suppose the sequence is 1, 1/2, 1/3, ... which converges to 0. The sine sequence is sin 1, sin 2, sin 3, ... which I highly doubt converges to 0.

 

[Unassuming]

The problem states that "if" x_n approaches 0 then the limit of (x_n)sin(1/x_n) approaches 0. I assume we have to use that information.

 

[gabbagabbahey]

Hint: try to show that -abs(x_n) \leq x_nsin \left(\frac{1}{x_n} \right) \leq abs(x_n) for all n. If you can do that, then just apply the squeeze theorem and your done.

Edit- you can't just say that lim x_n ->0 of x_nsin(1/x_n) is zero because x_n goes to zero...the limit of sin(1/x_n) does not exist so \lim ( x_n sin(1/x_n)) \neq (\lim (x_n))(\lim (sin(1/x_n)))

 

[Unassuming]

Please let me know if I need more rigor on this...

"Proof"

Since sin\frac{1}{x_n} is always between -1 and 1,

x_nsin\frac{1}{x_n} \leq x_n

Using similar logic,

x_n \leq x_nsin\frac{1}{x_n} \leq x_n

Since both negative and positive x_n \rightarrow 0

we know by the squeeze theorem that

x_nsin\frac{1}{x_n} \rightarrow 0

EDIT: How do I incorporate the abs?

 

[gabbagabbahey]

abs means absolute value (x_n could be negative for some n, so you must take the absolute value...suppose that there were a value of n for which x_n=-2 and sin(1/x_n)=+0.5, clearly x_n sin(1/x_n)=-1 which is not less that x_n=-2, but it is less than |-2|=2 )

\Rightarrow -|x_n| \leq x_nsin\frac{1}{x_n} \leq |x_n|

not

-x_n \leq x_nsin\frac{1}{x_n} \leq x_n

If x_n ->0, then so do Abs(x_n) and -Abs(x_n)

 

[Unassuming]

Okay, so here is what I have...

Proof

Since

\sin\frac{1}{x_n}

is always between -1 and 1,

x_n\sin\frac{1}{x_n} \leq |x_n|

Using similar logic,

-|x_n| \leq x_n\sin\frac{1}{x_n} \leq |x_n|

Since

-|x_n| \to 0

and

|x_n| \to 0

, we know by the squeeze theorem that

x_n\sin\frac{1}{x_n} \rightarrow 0