Limit of a Greatest Integer Function using Squeeze help

[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.

[lanedance]

what do the double brackets mean?

[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

[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 <=?

[lanedance]

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

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


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


-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 <=?

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

[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

[Kindayr]

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

and move from there?

[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