# Limit of a Greatest Integer Function using Squeeze help

1. Oct 23, 2009

### Kindayr

my midterm is in 4 hours and this actually the only thing i need help with.

1. The problem statement, all variables and given/known data
prove using squeeze theorem that lim(x-> +inf) (x^2 - [[x^2]])/x = 0

2. Relevant equations
g(x)<=f(x)<=h(x) [squeeze theorem]

3. The attempt at a solution
on the assignment i didn't know we had to use squeeze, so i just plugged in +inf and got 0, but we had to use squeeze. i wasn't there for his explanation of it.

all i need help with is setting up the inequality for the squeeze theorem and i'm fine, i'm just drawing blanks for all of this. help would be sooooo amazing. again, this is the only thing i'm confused on for my midterm.

2. Oct 23, 2009

### lanedance

what do the double brackets mean?

3. Oct 23, 2009

### Kindayr

the greatest integer function/the floor function.

so if its 5.5, the [[]] make it 5, if its -6.3 the [[]] make it -7

4. Oct 23, 2009

### Kindayr

i've been trying, and the only result i can get, and i don't know if its true is:

x-1 < [[x]] <= x

(x^2 - (x^2 - 1))/x < (x^2 - [[x^2]])/x <= (x^2 - (x^2))/x

-1/x < f(x) <= 0/x

-1/x < f(x) <= 0

(lim x-->+inf) -1/x = 0
(lim x-->+inf) 0 = 0

.:. through squeeze theorem, (lim x-->+inf) f(x) = 0

but does that squeeze theorem work if the inequality on the left is just < and not <=?

5. Oct 23, 2009

### lanedance

close, few mistakes though
first f is always positive, consider the squeeze limts you get, they don't really make sense

few changes here, first start with the assumption x>1 to reduce confusion later
the first line is correct
x-1< [[x]] <= x

now square everything
(x-1)2< [[x]]2 <= x2

multiply by -1, reversing the order
-x2<= -[[x]]2 < -(x-1)2

add x2 and divide by x gives

0=(x2 - x2)/x<= (x2-[[x]]2)/x < (x2-(x-1)2)/x

no its fine (if you use the correct squeeze as above), its still squeezed between 2 functions that go to the same limit

Last edited: Oct 23, 2009
6. Oct 23, 2009

### Kindayr

if you do it your way though:

x-1 < [[x]] <= x

(x-1)2 < [[x]]2 <= x2

first of all, does [[x2]] = [[x]]2?

then we goto:

-x2 <= -[[x]]2 < -(x-1)2

0/x <= f(x) < (x2 - x2 + 2x + 1)/x

don't we?

then you'd have 0 <= f(x) < 2

7. Oct 23, 2009

### Kindayr

should i start off with:
for x>1
x^2 - 1 < [[x^2]] < x^2

and move from there?

8. Oct 23, 2009

### lanedance

yeah that looks good, i misplaced the bracket,

so for x>1
x - 1 < [[x]] <= x
giving
x2 - 1 < [[x2]] <= x2
minus
- x2 < -[[x2]] <= -x2+ 1
then +x2
0 = x2- x2 <x2 -[[x2]] <= x2-x2+ 1 = 1
then /x

0/x< (x2 -[[x2]])x <= 1/x as required