Compute this limit please:?

Question

http://m1.freeshare.us/155fs15152.gif

[x] represnts integer part of x.

Answer 1

If x is an integer, then the limit is x/(x-1). Otherwise, the limit is 0. To understand this, let x = z + a, where z is the integer part, and a is the decimal part. Then we have:

Limit(n->∞) (z(z^n-1) / (z-1)(z+a)^n), or

Limit(n->∞) ((z/(z-1))(z/(z+a))^n - (z/(z-1))(z+a)^n

The right term drops out, leaving the left term having either the limit value of z/(z-1) if a = 0, and 0 if a ≠ 0.

Answer 2

I think the catch is that the power of n is *inside* the brackets. So it's not actually a geometric series.

The last term is [x^n]/x^n. Since the difference between [x^n] and x^n is necessarily less than 1 for every n, the ratio [x^n]/x^n approaches 1.

Ignoring all other terms for large n, I think the limit will be 1.

If the last term had been [x]^n/x^n, then the limit would have been 0.

Answer 3

I may be wrong, but I found the answer to be zero.
For each n, I viewed the ratio as a geometric series,
with r = [x] and worked out a general Sn.
I starred your question, so I will leave it to others to verify.
EDIT: why math prof? [x] is less than x so [x]^n / x^n will go to zero.
Here is a more detailed solution from me and I might be wrong.

[x]/ x^n + [x]^2/ x^n + [x]^3/ x^n + ...[x]^n/ x^n.....for each n
this is a GP, with r = [x]
Sn = [x]( [x]^n - 1) /[( x^ n)( [x] - 1)] =
[x]/ ([x] -1) * ( [x]^n - 1 / x^n )
( [x]^n - 1 / x^n ) = [x]^n/ x^n - 1/x^n and [x]/ x < 1 so
[x]^n/x^n will go to zero and Sn will go to zero as n goes to infinity. provided that x is not an integer.if it is, it will diverge
Please let me know if there is an error in my logic.
ty free variable, i did not pay attention to that. I think you are right., the limit is 1.

Answer 4

The only term that doesn't go to zero as n tends to infinity is the last term, which goes to one.
My two cents is that the limit is one.

Answer 5

∞