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what about them?

etc

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

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

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

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

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

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

@ospreytriple that is what i think as well

So the graph is nor a normal curve?

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

|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 ?

floor is a step function

Right right.

what is left to do?

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

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

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!!