## anonymous one year ago another question about limit.

1. anonymous

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

2. jtvatsim

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

3. anonymous

yes .

4. jtvatsim

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

5. anonymous

@ganeshie8 can you help me?

6. anonymous

@iambatman

7. 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

8. Astrophysics

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

9. jtvatsim

Brilliant @astrophysics and @ganeshie8 ! :D

10. ganeshie8

both xcotx-x and xcotx have same limits

11. ganeshie8

12. Astrophysics

xD

13. jtvatsim

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

14. ganeshie8

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

15. jtvatsim

16. ganeshie8

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

17. jtvatsim

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

18. 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

19. jtvatsim

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

20. 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

21. jtvatsim

agreed with above reasoning^^

22. 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

23. ganeshie8

thnks, just convincing myself.. :)

24. Astrophysics

Lol nice you really squeezed that one through

25. Astrophysics

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

26. ganeshie8

yeah [x]cotx is tricky

27. 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 $\lim_{x \rightarrow 0^{+}}(x - x)\cot x =0$

28. 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!

29. Astrophysics

Amazing job from both of you! Night! :p

30. ganeshie8

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