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

[Jbreezy]

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

Homework Statement

Quick question. Brain flagrance.

Homework Equations

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

The Attempt at a Solution

?? Why is this. How do you know.

 

[Mark44]

Homework Statement

Quick question. Brain flagrance.

Homework Equations

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

The Attempt at a Solution

?? Why is this. How do you know.

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

 

[Jbreezy]

Got it. What about this? THanks
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.

 

[Jbreezy]

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

 

[Mark44]

Got it. What about this? THanks
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.

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.

 

[Mark44]

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

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

 

[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?

 

[Mark44]

Sorry, I was typing faster than I was thinking. I meant "what does u do" as n → ∞?(I edited my earlier post.)

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?

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.

 

[reinloch]

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.

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