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

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