How Can I Prove the Limit of a Product Equals Zero with an Undefined Limit?

[ptolema]

Homework Statement

the problem is: prove that if lim x-->0 g(x)=0, then lim x-->0 g(x)sin(1/x)=0. it seems simple enough, but certain parts have me wondering.

Homework Equations

so i already know that lim x-->0 sin(1/x) is undefined because the function oscillates near 0. I'm not really sure how to incorporate that without just stating it, though. also, how do i show the effect of lim x-->0 g(x)=0? if a limit is undefined, does it not exist?

The Attempt at a Solution

i'm not sure if this is completely erroneous but my tentative solution is the following:
if lim x-->0 g(x)=0, lim x-->0 g(x)sin(1/x) = [lim x-->0 g(x)]*[lim x-->0 sin(1/x)]= 0*lim x-->0 sin(1/x)
since a number times 0 equals 0, it follows that lim x-->0 g(x)sin(1/x) = 0.

please help!

 

[slider142]

i'm not sure if this is completely erroneous but my tentative solution is the following:
if lim x-->0 g(x)=0, lim x-->0 g(x)sin(1/x) = [lim x-->0 g(x)]*[lim x-->0 sin(1/x)]= 0*lim x-->0 sin(1/x)
since a number times 0 equals 0, it follows that lim x-->0 g(x)sin(1/x) = 0.

This argument is erroneous since $\lim_{x\rightarrow 0} \sin \frac{1}{x}$ is not a number, and furthermore the limit of the product is not the product of the limits of the factors unless the limits of the factors both exist.
Since the second limit does not exist, you will have to perform an epsilon-delta proof of the statement, or use a theorem about limits of bounded functions.

 

[ptolema]

would i start off showing the epsilon-delta proof of lim x-->0 g(x)=0, or of lim x-->0 g(x)sin(1/x) = 0? since lim x-->0 sin(1/x) DNE, it doesn't have an applicable epsilon-delta form, so how can i include it in an epsilon-delta proof?

 

[Gavins]

Have you tried the squeeze theorem?

 

[ptolema]

how would squeeze theorem apply? i don't know if g(x)$$\leq$$g(x)sin(1/x) or any of the other conditions for squeeze theorem.

edit:
ah, i see now. squeeze theorem can be used because i know the range of the sine function. that definitely clears things up, as i already proved the theorem earlier. thanks for the help!

 

[Gavins]

The sine function is bounded.
$$-1 \leq sin(1/x) \leq 1$$

Now incorporate g(x).