How do I find the limit as n approaches infinity of a fraction raised to the nth degree. Ex: (3/4)^n?

Question

lim (3/4)^n n->inf

I attempted to find the limits of 3^n and 4^n separately, but I find myself with the indeterminate form inf/inf.

Should l'hopital's rule be used?

Answer 1

Think about it a little bit. For (3/4)^n, the number is getting smaller as n gets larger. Since every term is positive, there *is* a limit. I L is the limit, however, L=L*(3/4), so L=0. This works for x^n whenever 0<=x<1. Clearly 1^n=1. If x>1, then x^n gets larger as n gets larger. By the previous reasoning, if there were a limit, the limit would have to be 0. Thus, there is no limit. Thus we have:
lim x^n
= 0 if -1 =1 if x=1
is infinite if x>1
fails to exist if x<=-1.

Answer 2

In this case, zero. The thing you need to ask is "is the fraction more or less than 1?" If the fraction is greater than 1, the limit is infinity, if it is exactly 1, the limit is 1, and if the fraction isnonnegative and less than 1, the limit is zero.

Answer 3

you are able to't "cancel out" infinity contained in the denominator and nominator of the fraction. Sorry. Math is a precise technological expertise. What you may do is to cancel out aspect n^100 to blame for the boom: as long as (n+30)^100 / (2 * n^100)=0.5*(n^100/n^100)* *(a million+30/n)^100 =0.5*(a million+30/n)^100 and 30/n->0 as n->infinity, then 0.5*(a million+30/n)^100 ->0.5*a million=0.5=a million/2 lim (n+30)^100 / (2 * n^100)=a million/2.

Answer 4

if you're raising a proper fraction to n, then the limit as n approaches infinity is 0.

if you're raising an improper fraction to n, then the limit as n approaches infinity is undefined.

Answer 5

in a case like this.. which one approaches infinity faster? 3^n or 4^n?... then there are rules for determining some other functions

Answer 6

Let y = the limit as x --> infinity of b^x.
ln[y] = ln[lim as x--> inf of b^x)
Since ln[x] is continuous over its domain,
ln[y] = lim as x-->inf of ln[b^x] = lim as x-->inf of xln[b]

If b is greater than or equal to 0:
If b < 1 ln[b] < 0 and lim as x-->inf of xln[b] = -inf, so y = 0.
If b = 1 ln[b] = 0 and lim as x-->inf of xln[b] = 0, so y = e^0 = 1.
If b > 1 ln[b] > 0 and lim as x-->inf of xln[b] = +inf, so y = +inf.

Answer 7

Damn, that limit stuff sucked. Just graph it into your calculator (if you've got a TI-82 or higher). it would be
y=(3/4)^x

I don't remember. Oh, early_sol's right.