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anonymous
 one year ago
another question about limit.
anonymous
 one year ago
another question about limit.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0^{+}}(x\left[cotx \right]+\left[ x \right]cotx)\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Are those greatest integer functions? For the [ ] brackets?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1OK, so we have rounding here like [3.141] = 3. Got it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 can you help me?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Consider below limit first : \[\lim_{x \rightarrow 0^{+}}(x\left[\cot x \right])\] By definition of floor function we have \[\cot x1\lt [\cot x] \le \cot x\] Multiply \(x\) through out and get \[x(\cot x1)\lt x[\cot x] \le x\cot x\] Take limit through out and use squeeze thm. You can work the other limit similarly

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha I just suggested squeeze theorem to jtvatsim xD asking if it was related

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Brilliant @astrophysics and @ganeshie8 ! :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4both xcotxx and xcotx have same limits

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you 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

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Ah right, @ganeshie8 ... good thinking. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4that strict inequality on left hand side.. is that a problem for squeezing ?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1hmm... not positive about that...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4same here.. need to think it through...

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1perhaps it's fine because of the x > 0+ from the right???

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4wolf 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
 one year ago
Best ResponseYou've already chosen the best response.1I really hope I am wrong, but it appears that we have a problem with using the squeeze theorem on the second limit....

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I 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
 one year ago
Best ResponseYou've already chosen the best response.1agreed with above reasoning^^

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4f(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
 one year ago
Best ResponseYou've already chosen the best response.4thnks, just convincing myself.. :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Lol nice you really squeezed that one through

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I think it does work out, as you said they both = 1 right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4yeah [x]cotx is tricky

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4does 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
 one year ago
Best ResponseYou've already chosen the best response.1Yep, exactly what I just put in my pdf. lol :) I'm attaching it here and checking out... it's really late here. Nights!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Amazing job from both of you! Night! :p

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Nice :) kinda same idea but it looks much better with the correct steps and reasons! gnite!
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