another question about limit.

- anonymous

another question about limit.

- katieb

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- anonymous

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

- jtvatsim

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

- anonymous

yes .

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## More answers

- jtvatsim

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

- anonymous

@ganeshie8 can you help me?

- anonymous

- ganeshie8

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

- Astrophysics

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

- jtvatsim

Brilliant @astrophysics and @ganeshie8 ! :D

- ganeshie8

both xcotx-x and xcotx have same limits

- ganeshie8

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

- Astrophysics

xD

- jtvatsim

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

- ganeshie8

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

- jtvatsim

hmm... not positive about that...

- ganeshie8

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

- jtvatsim

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

- ganeshie8

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

- jtvatsim

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

- ganeshie8

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 :
|dw:1439538710111:dw|
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

- jtvatsim

agreed with above reasoning^^

- ganeshie8

f(x) < g(x) < h(x)
so if the limits f(x) and h(x) are equal, then it follows that g(x) has the same limit

- ganeshie8

thnks, just convincing myself.. :)

- Astrophysics

Lol nice you really squeezed that one through

- Astrophysics

I think it does work out, as you said they both = 1 right?

- ganeshie8

yeah [x]cotx is tricky

- ganeshie8

does this look okay for second term
Let \(\{x\}\) represent the fractional(decimal) part of \(x\), then \([x] = x-\{x\}\)
\[\lim_{x \rightarrow 0^{+}}\left[ x \right]\cot x = \lim_{x \rightarrow 0^{+}}(x -\{x\})\cot x \]
Since for \(x\in (0,1)\) we have \(x=\{x\}\), that becomes
\[\lim_{x \rightarrow 0^{+}}(x - x)\cot x =0\]

- jtvatsim

Yep, exactly what I just put in my pdf. lol :) I'm attaching it here and checking out... it's really late here. Nights!

##### 1 Attachment

- Astrophysics

Amazing job from both of you! Night! :p

- ganeshie8

Nice :) kinda same idea but it looks much better with the correct steps and reasons! gnite!

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