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...
Printable View
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...
It is completely transparent thatQuote:
Originally Posted by Mush89
gets very very large as
.
What does that tell us about
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...
NEVER, NEVER write that. It is an automatic F.Quote:
Originally Posted by Mush89
Now are you asked to prove the limit is 0?
Or are you simply asked to find the limit?
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."
I truly that you were not actually taught that.Quote:
Originally Posted by Mush89
But then again there are a lot of calculus teachers that have no business being in any mathematics class room.
I expect you were told thatQuote:
Originally Posted by Mush89
. You can NOT ever substitute
as though it's a number. It's more of an idea.