Squeeze Theorem greatest integer function

- Babynini

Squeeze Theorem greatest integer function

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- Babynini

##### 1 Attachment

- anonymous

what about them?

- 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

then on \(-\frac{\pi}{2}\) to \(0\) it goes from \(0\) to \(1\) so on that interval it would be \(0\)

- anonymous

etc

- Empty

Might be fun to play here, the greatest integer function is sometimes called the ceiling function. https://www.desmos.com/calculator/fdipg03oxv

- anonymous

Do those funky brackets mean the least integer?

- anonymous

on the interval \(0\) to \(\frac{\pi}{2}\) it goes from \(1\) to \(0\) so it is 0 there as well

- Babynini

So if I were to draw one cosine it would be just one mountain

- anonymous

@empty i though greatest integer was floor

- anonymous

greatest integer less than
like \(\lfloor 2.4\rfloor=2\)

- Babynini

|dw:1444265853694:dw|
like that!..? haha i'm such an artist.

- anonymous

Why can't they use the standard \[\lceil \cos x \rceil\] for ceiling and \[\lfloor \cos x \rfloor\]for floor?

- anonymous

who said art was dead
that is cosine though not \(\lfloor \cos(x)\rfloor\)

- anonymous

"greatest integer" to me means "greatest integer less than" so "floor"
whereas "least integer" means "least integer greater than i.e. "ceiling"

- 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

Wait, so how is cos(x) looking then? o.o

- Empty

Almost identical but shifted https://www.desmos.com/calculator/cowtkonbwe

- Babynini

hmm..shifted how? Like the only difference seems to be that there's lines across the bottom and top.

- anonymous

it is identically \(-1\) on \((-\pi,-\frac{\pi}{2})\)

- anonymous

then it is \(0\) on \(-\frac{\pi}{2},\frac{\pi}{2}\) except it is 1 at \(0\)

- anonymous

then \(-1\) again

- anonymous

http://www.wolframalpha.com/input/?i=floor%28cos%28x%29%29%2C+-pi..pi

- anonymous

you can't see the \((0,1)\) on wolfram, but it is there

- 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

or here
https://www.desmos.com/calculator/cowtkonbwe

- anonymous

@ospreytriple that is what i think as well

- 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

So the graph is nor a normal curve?

- Babynini

Is it just a point at (0,1)?

- 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

|dw:1444266645781:dw|

- Babynini

Is what ..i'm imagining. By what yall are saying 0.0

- Empty

Yeah that's right, it's what we're saying. :)

- Babynini

|dw:1444266755110:dw|
like dis?

- Babynini

The line on 0 being all but the point (0,0)

- Empty

Yeah, since the greatest integer function says [[0]]=0 Looks good to me.

- Babynini

Fabulous! phew. part b!

- Babynini

i) = 0 ?

- 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

@SolomonZelman

- anonymous

the reason that desmos thing threw you off is that they graphed both the floor function and cosine on the same graph

- anonymous

floor is a step function

- Babynini

Right right.

- anonymous

what is left to do?

- 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

Part b and c.

- anonymous

\[\lim_{x\to 0}f(x)=0\] for sure, since it is identically 0 to the right and to the left

- Babynini

b:
i) 0
ii) if the rules apply then.. 0.57 ?

- anonymous

the fact that \(f(0)=1\) does not effect the limit

- anonymous

what does "rules apply" mean?
this thing never approaches 0.57
it is either 0 or -1

- 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

\[\lim_{x\to \frac{\pi}{2}}f(x)=0\]

- anonymous

ok typo there hold on

- anonymous

\[\huge \lim_{x\to \frac{\pi}{2}^-}f(x)=0\]

- anonymous

since it is identically 0 on the interval to the left

- Babynini

They all look like they're ultimately approaching 0

- anonymous

no, from the right it is \(-1\)

- anonymous

|dw:1444267958342:dw|

- anonymous

\[\huge \lim_{x\to \frac{\pi}{2}^+}f(x)=-1\]

- anonymous

and since the right and left hand limits are not the same, the limit does not exist

- Babynini

Righto, gotcha.

- anonymous

k good
to answer the last question, the limit exists everywhere except at those two jumps

- Babynini

So I should probably write that in some form of interval haha the two jumps meaning.

- anonymous

no limit at \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\)

- anonymous

at those to points it jumps form -1 to 0 and from 0 to -1 receptively

- anonymous

|dw:1444268282111:dw|

- Babynini

hahah awesome. Art lives!!

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