Question

Limit question.

Answer 1

\[\Huge \lim_{x \rightarrow \frac{\pi}{2}} [\frac{x-\frac{\pi}{2}}{\cos x}]\]
where [.] is the greatest integer

Answer 2

Recall that\[-\cos x=\sin\left(x-\frac{\pi}{2}\right)\]

Answer 3

yep.

Answer 4

Right, so you might consider the substitution \(y=x-\dfrac{\pi}{2}\), so that \(y\to0\) as \(x\to\dfrac{\pi}{2}\), so the limit is equivalent to
\[\lim_{y\to0}\left[\frac{y}{-\sin y}\right]\]
Does this look familiar?

Answer 5

limit y->0 siny/y or y/siny is 1 so what would be the greatest integer of it?

Answer 6

Right, don't forget the negative sign.

Answer 7

so answer is -1?

Answer 8

Yes

Answer 9

http://www.wolframalpha.com/input/?i=limit+x-%3E+pi%2F2+floor%28%28x-pi%2F2%29%2Fcosx%29%29

Answer 10

this is what has been bugging me for the past hour

Answer 11

Oh I see, I was making the mistake of thinking \(\lfloor-x\rfloor=-\lfloor x\rfloor\), whereas it should actually be \(\lfloor-x\rfloor=-\lceil x\rceil\). Sorry about that!

Answer 12

didn't you just write the same thing twice? :O

Answer 13

I mean you wrote [-x]=-[x] the thing only twice :P

Answer 14

Oh, I see what you mean. No, these are different functions - notice the little tails on the LHS are on the bottom of the brackets, whereas they're on the top on the RHS. This notation is used to differentiate between the floor and ceiling functions. The floor function is equivalent to the greatest integer function.

Answer 15

You can think of the ceiling function as "floor plus 1".

Answer 16

ohh my bad,thanks !:)

Answer 17

yw