what's the limit of the sequence, (3^(n+1))/(4^n-1))?

Question

n approaches infinity, duh

Answer 1

This is the same as 12*3^n/4^n.
So we really need to know what happens to 3^n/4^n as n approaches infinity.

If we look at n = 1 we get 3/4. For n=2 we get 9/16 and for n= 3 we get 27/64. Clearly each term is less than the previous term. It is also clear that 3^n/4^n will always be positive, that is > 0, so the limit must be zero.

If the terms of an infinite sequence steadily decrease butremain greater than som constant c, then the sequence has a limit and is convergent: this limit is either c or some nuber greater than c. In this case, the limit is c=0.

Answer 2

the limit is 0.
For n large, 4^n is the dominant term so the seuence behaves like 3^(n+1)/4^n = 3(3^n)/4^n
(3/4)^n is a geometric sequence with abs r < 1, it goes to 0. Hence limit is 0.
Okay, I have misread you denominator as (4^n) - 1, but the answer is the same. 0

Answer 3

lim n---> ?: ( 3n² + a million) / ( 4n² - 2n + 5) What you may want to do is divide every time period in right here with the help of n² lim n---> ?: ( 3 + a million/n²) / (4 - 2/n + 5/n²) Now plug in ? for all values of n. A person-friendly rule of math is that that in case you divide a small style with the help of an truly huge style the answer is going to be so small, that you'll in basic terms round all the way down to 0, which precisely what takes position right here. lim n---> ? ( 3 + 0) / ( 4 - 0 + 0) very last answer: 3 / 4

Answer 4

3^(n+1)=9*3^(n-1)
(3^(n+1))/(4^n-1)) =9*[(3/4)^(n-1)]
Since 3/4<1 The limit will be zero

Answer 5

there is no limit just out of nowhere, u have to specify the limit as n approaches what number. You need to review the definition of limits.