As n approaches infinity.

I tried using Squeeze Theorem but I got stuck...

I said -1 < 1 < 1

Then, -1/((2^n)) < 1/((2^n) < 1/((2^n) and I'm stuck...

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- Apr 5th 2011, 02:34 PMMush89What is the limit of 1/(2^(n))
As n approaches infinity.

I tried using Squeeze Theorem but I got stuck...

I said -1 < 1 < 1

Then, -1/((2^n)) < 1/((2^n) < 1/((2^n) and I'm stuck... - Apr 5th 2011, 02:40 PMPlato
- Apr 5th 2011, 02:43 PMMush89
1/(2^n) = 1/(infinity) = 0, but is it that simple? It's for a calc. class and this is the limit of a sequence...

- Apr 5th 2011, 02:49 PMPlato
- Apr 5th 2011, 02:53 PMMush89
I was taught that 1 is a constant, and that c/(infinity) = 0.

And it says, "Does the limit exist, and if so, what is it's value? Give detailed explanation." - Apr 5th 2011, 02:57 PMPlato
- Apr 5th 2011, 08:27 PMProve It