The discussion revolves around the limit of the expression (3/4)^n as n approaches infinity, questioning why this limit equals zero. The subject area includes concepts from calculus, specifically limits and sequences.
The discussion is ongoing, with participants raising questions about the reasoning behind limits and substitutions. Some guidance is offered regarding the relationship between n and the variable u in the context of limits, but no consensus has been reached.
Participants express confusion over the assumptions made in the problems, particularly regarding the behavior of sequences and the implications of constants in limits as n approaches infinity.
Jbreezy said: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.
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.Jbreezy said: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.
For the substitution u = k/n, if n → ∞, what does u do?Jbreezy said:Ps then after they make the sub with u they write it as u goes to 0...huh?
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.Jbreezy said: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 said: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.