Squeeze Theorem greatest integer function

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Squeeze Theorem greatest integer function

Mathematics
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what about them?
from \(-\pi \) to \(-\frac{\pi}{2}\) cosiene goes from \(-1\) to \(0\) so it would be identically \(-1\) on that interval

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then on \(-\frac{\pi}{2}\) to \(0\) it goes from \(0\) to \(1\) so on that interval it would be \(0\)
etc
Might be fun to play here, the greatest integer function is sometimes called the ceiling function. https://www.desmos.com/calculator/fdipg03oxv
Do those funky brackets mean the least integer?
on the interval \(0\) to \(\frac{\pi}{2}\) it goes from \(1\) to \(0\) so it is 0 there as well
So if I were to draw one cosine it would be just one mountain
@empty i though greatest integer was floor
greatest integer less than like \(\lfloor 2.4\rfloor=2\)
|dw:1444265853694:dw| like that!..? haha i'm such an artist.
Why can't they use the standard \[\lceil \cos x \rceil\] for ceiling and \[\lfloor \cos x \rfloor\]for floor?
who said art was dead that is cosine though not \(\lfloor \cos(x)\rfloor\)
"greatest integer" to me means "greatest integer less than" so "floor" whereas "least integer" means "least integer greater than i.e. "ceiling"
Oh floor ceil, whatever haha yeah I agree with @ospreytriple I wouldn't have messed up if they just wrote \[\left\lfloor \frac{1}{\sec x} \right\rfloor\]
Wait, so how is cos(x) looking then? o.o
Almost identical but shifted https://www.desmos.com/calculator/cowtkonbwe
hmm..shifted how? Like the only difference seems to be that there's lines across the bottom and top.
it is identically \(-1\) on \((-\pi,-\frac{\pi}{2})\)
then it is \(0\) on \(-\frac{\pi}{2},\frac{\pi}{2}\) except it is 1 at \(0\)
then \(-1\) again
http://www.wolframalpha.com/input/?i=floor%28cos%28x%29%29%2C+-pi..pi
you can't see the \((0,1)\) on wolfram, but it is there
There seems to be some confusion above somewhere. The greatest integer function is also called the floor function, the greatest integer that is less than or equal to the argument.
or here https://www.desmos.com/calculator/cowtkonbwe
@ospreytriple that is what i think as well
So there's no..normal curve. Ah yeah @Empty showed me the desmos one but I don't get it really haha it looks like a cosine but with lines on the hills/indents
So the graph is nor a normal curve?
Is it just a point at (0,1)?
That's definitely it @satellite73 . The ceiling function mentioned above is also called the least integer function, the smallest integer that is greater than or equal to the argument.
|dw:1444266645781:dw|
Is what ..i'm imagining. By what yall are saying 0.0
Yeah that's right, it's what we're saying. :)
|dw:1444266755110:dw| like dis?
The line on 0 being all but the point (0,0)
Yeah, since the greatest integer function says [[0]]=0 Looks good to me.
Fabulous! phew. part b!
i) = 0 ?
So for this one, the rules we used for the previous problem (ie: a, or a-1) do not apply? or do they?
the reason that desmos thing threw you off is that they graphed both the floor function and cosine on the same graph
floor is a step function
Right right.
what is left to do?
I'm wondering about the part b now. Because I know the rules, which you stated for the last problem. But I wasn't sure if those applied here.
Part b and c.
\[\lim_{x\to 0}f(x)=0\] for sure, since it is identically 0 to the right and to the left
b: i) 0 ii) if the rules apply then.. 0.57 ?
the fact that \(f(0)=1\) does not effect the limit
what does "rules apply" mean? this thing never approaches 0.57 it is either 0 or -1
the rules of if it approaches from the left it = a - 1 and if it approaches from the right it = a but for this one I guess we just use the graph.
\[\lim_{x\to \frac{\pi}{2}}f(x)=0\]
ok typo there hold on
\[\huge \lim_{x\to \frac{\pi}{2}^-}f(x)=0\]
since it is identically 0 on the interval to the left
They all look like they're ultimately approaching 0
no, from the right it is \(-1\)
|dw:1444267958342:dw|
\[\huge \lim_{x\to \frac{\pi}{2}^+}f(x)=-1\]
and since the right and left hand limits are not the same, the limit does not exist
Righto, gotcha.
k good to answer the last question, the limit exists everywhere except at those two jumps
So I should probably write that in some form of interval haha the two jumps meaning.
no limit at \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\)
at those to points it jumps form -1 to 0 and from 0 to -1 receptively
|dw:1444268282111:dw|
hahah awesome. Art lives!!

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