# Thread: What is the limit of 1/(2^(n))

1. ## What 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...

2. Originally Posted by Mush89
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 that $\displaystyle 2^n$ gets very very large as $\displaystyle n\to\infty$.
What does that tell us about $\displaystyle \dfrac{1}{2^n}~?$

3. 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...

4. Originally Posted by Mush89
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.
Now are you asked to prove the limit is 0?
Or are you simply asked to find the limit?

5. 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."

6. Originally Posted by Mush89
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.
But then again there are a lot of calculus teachers that have no business being in any mathematics class room.

7. Originally Posted by Mush89
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 expect you were told that $\displaystyle \lim_{x \to \infty}\frac{c}{x} = 0$. You can NOT ever substitute $\displaystyle \displaystyle \infty$ as though it's a number. It's more of an idea.