Question

another question about limit.

Answer 1

\[\lim_{x \rightarrow 0^{+}}(x\left[cotx \right]+\left[ x \right]cotx)\]

Answer 2

Are those greatest integer functions? For the [ ] brackets?

Answer 3

yes .

Answer 4

OK, so we have rounding here like [3.141] = 3. Got it.

Answer 5

@ganeshie8 can you help me?

Answer 6

@iambatman

Answer 7

Consider below limit first :

\[\lim_{x \rightarrow 0^{+}}(x\left[\cot x \right])\]

By definition of floor function we have \[\cot x-1\lt [\cot x] \le \cot x\]
Multiply \(x\) through out and get \[x(\cot x-1)\lt x[\cot x] \le x\cot x\]

Take limit through out and use squeeze thm.
You can work the other limit similarly

Answer 8

Haha I just suggested squeeze theorem to jtvatsim xD asking if it was related

Answer 9

Brilliant @astrophysics and @ganeshie8 ! :D

Answer 10

both xcotx-x and xcotx have same limits

Answer 11

you see now.. when you think something in ur head, your ideas do float around w/o your permission sometimes... and others get access to them ;p

@Astrophysics

Answer 12

xD

Answer 13

Ah right, @ganeshie8 ... good thinking. :)

Answer 14

that strict inequality on left hand side.. is that a problem for squeezing ?

Answer 15

hmm... not positive about that...

Answer 16

same here.. need to think it through...

Answer 17

perhaps it's fine because of the x --> 0+ from the right???

Answer 18

wolf says both sides limit exist and equal to 1
http://www.wolframalpha.com/input/?i=%5Clim_%7Bx+%5Crightarrow+0%7D+%28x*floor%28cotx%29%29

Answer 19

I really hope I am wrong, but it appears that we have a problem with using the squeeze theorem on the second limit....

Answer 20

I think that strict inequality is not an issue because we don't bother about the value of function at \(x=0\), we're only interested in what goes on around :




It doesn't matter whether the functions are defined at \(x=0\) or if their values all are "equal"... so squeeze thm should work just fine with strict inequalities too