Why Doesn't the Squeeze Theorem Apply to lim_{x\rightarrow0}\frac{sin(x)}{x}?

[camcool21]

I just read through a geometric proof that made complete sense to me and resulted in:

$$lim_{x\rightarrow0}\frac{sin(x)}{x} = 1$$

But this contradicts what I thought I knew about the squeeze/sandwich principle.

-1 < sin(x) < 1

(-1/x) < [sin(x)/x] < (1/x)

The limits at (-1/x) and (1/x) don't exist when x is zero, because the left and right hand limits are positive and negative infinity for both. Why can the limit of the term in the middle be anything other than non-existent according to the squeeze theorem? I know it must equal 1, due to the geometric proof I just read, but now I'm just confused.

 

[eczeno]

the squeeze theorem can only be applied when both of the 'squeezing' limits exist. As you correctly point out, neither of the ones you are using exist, so your example does not say anything about the squeeze theorem. as for proving your limit, convince yourself that:

cos(x) < sin(x)/x < 1 for -pi/2 < x < pi/2

and, well, squeeze.

 

[camcool21]

Thanks for the reply. I suppose that makes sense. If the either of the two "squeezing" functions don't have a limit, it is impossible to say anything about the middle function with the squeeze theorem.

The fact that cos(x) < sin(x)/x < 1 for -pi/2 < x < pi/2, was the result of the geometric proof I had mentioned. It's in Adrian Banner's "The Calculus Lifesaver" and it's pretty nifty. It involves comparing different areas in a sliced up chunk of the unit circle.