# Homework Help: Why limit n-> infinity (3/4)^n = 0

1. Jul 1, 2013

### Jbreezy

Why limit n--> infinity (3/4)^n = 0

1. The problem statement, all variables and given/known data

Quick question. Brain flagrance.

2. Relevant equations

Why limit n--> infinity (3/4)^n = 0

3. The attempt at a solution

?? Why is this. How do you know.

2. Jul 1, 2013

### Staff: Mentor

Look at the sequence 3/4, (3/4)2, (3/4)3, and so on. What appears to be happening?

3. Jul 1, 2013

### Jbreezy

limit (1+k/n)^n as n->infinity

Just go here http://www.calcchat.com/book/Calculus-ETF-5e/
You have to put in chapter 9 section 1 and question 69. I just don't understand how they knew to make a substitution.

4. Jul 1, 2013

### Jbreezy

Ps then after they make the sub with u they write it as u goes to 0....huh?

5. Jul 1, 2013

### Staff: Mentor

They are likely using a result from a previous problem. If so, the purpose of the substitution was to get the problem in that form.

6. Jul 1, 2013

### Staff: Mentor

For the substitution u = k/n, if n → ∞, what does u do?

Last edited: Jul 2, 2013
7. Jul 1, 2013

### Jbreezy

Stays the same? Isn't k a constant? So OH lol so you take the limit as u goes to 0 because as k/n goes to infinity u goes to 0? I say that right?

8. Jul 2, 2013

### Staff: Mentor

Sorry, I was typing faster than I was thinking. I meant "what does u do" as n → ∞?(I edited my earlier post.)
No, you didn't. As k/n goes to infinity, so does u, since u = k/n. The way to look at it is, as n → ∞, then the fraction k/n → 0.

9. Jul 3, 2013

### reinloch

This always bugs me, because assuming k > 0, as n → ∞, k/n → 0 from the right.