anonymous
  • anonymous
another question about limit.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\lim_{x \rightarrow 0^{+}}(x\left[cotx \right]+\left[ x \right]cotx)\]
jtvatsim
  • jtvatsim
Are those greatest integer functions? For the [ ] brackets?
anonymous
  • anonymous
yes .

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jtvatsim
  • jtvatsim
OK, so we have rounding here like [3.141] = 3. Got it.
anonymous
  • anonymous
@ganeshie8 can you help me?
anonymous
  • anonymous
ganeshie8
  • 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
  • Astrophysics
Haha I just suggested squeeze theorem to jtvatsim xD asking if it was related
jtvatsim
  • jtvatsim
Brilliant @astrophysics and @ganeshie8 ! :D
ganeshie8
  • ganeshie8
both xcotx-x and xcotx have same limits
ganeshie8
  • 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
  • Astrophysics
xD
jtvatsim
  • jtvatsim
Ah right, @ganeshie8 ... good thinking. :)
ganeshie8
  • ganeshie8
that strict inequality on left hand side.. is that a problem for squeezing ?
jtvatsim
  • jtvatsim
hmm... not positive about that...
ganeshie8
  • ganeshie8
same here.. need to think it through...
jtvatsim
  • jtvatsim
perhaps it's fine because of the x --> 0+ from the right???
ganeshie8
  • 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
  • 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
  • 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
  • jtvatsim
agreed with above reasoning^^
ganeshie8
  • 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
  • ganeshie8
thnks, just convincing myself.. :)
Astrophysics
  • Astrophysics
Lol nice you really squeezed that one through
Astrophysics
  • Astrophysics
I think it does work out, as you said they both = 1 right?
ganeshie8
  • ganeshie8
yeah [x]cotx is tricky
ganeshie8
  • 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
  • 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!
Astrophysics
  • Astrophysics
Amazing job from both of you! Night! :p
ganeshie8
  • ganeshie8
Nice :) kinda same idea but it looks much better with the correct steps and reasons! gnite!

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