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Limits and derivatives

  1. May 20, 2016 #1
    1. The problem statement, all variables and given/known data
    What will be lim[2sin(x-1)/(x-1)], where x tends to 1?
    [ ] denotes greatest integer function.
    2. Relevant equations
    Can I directly solve it using the formula sinx/x =1 when x tends to 0

    3. The attempt at a solution
    Okay so the quantity inside [ ] can be written as ——>>2 sin(x-1)/(x-1).. And sin(x-1)/(x-1) should equAte to 1 and only 2 should be left inside the greAtest integer function and therefore the answer should be 2... But the answer is 1 what am I doing wrong please explain...
     
  2. jcsd
  3. May 20, 2016 #2
    You really should be right. Can you make sure you wrote it down correctly?
     
  4. May 20, 2016 #3
    Lol yes it the same question I too was shell shocked to see the answer
     
  5. May 20, 2016 #4

    Samy_A

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    What you claim is that ##\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)] = [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]##.

    But is that claim correct? Is [] a continous function?
     
  6. May 20, 2016 #5

    SammyS

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    This limit is not 2 .

    The given answer, that the limit is 1, is indeed correct.
     
  7. May 20, 2016 #6
    How did you get to that? Could you please explain?
     
  8. May 20, 2016 #7

    Samy_A

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    Can you explain why you think that ##\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)] = [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]##?
     
  9. May 20, 2016 #8
    I don't know what effect does it really have? So please explain it to me in detail
     
  10. May 20, 2016 #9

    Samy_A

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    Let's start with the RHS: ##\displaystyle \ [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]##.
    As you stated, ##\displaystyle \lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))=2##, so that ##\displaystyle \ [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]=[2]=2##.

    For the LHS, ##\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)]##, you can't simply reverse the order of operations, swapping the limit with the []. That only works for continuous functions.

    [] is not continuous at x=1. If that is not clear to you, try to prove this first.

    To evaluate ##\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)]##, you first have to evaluate ##[2\sin(x-1)/(x-1)]## for x close to 1, but not equal to 1. Try it, and see what you get.
     
  11. May 20, 2016 #10
    I have no idea what you're telling me
     
  12. May 20, 2016 #11

    Samy_A

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    That was quick.

    Anyway, then we are back at post #7. Why did you think that the limit should be 2? Give your detailled reasoning, then someone will hopefully be able to help you find the error(s).
    Because, as @SammyS wrote, the correct limit is 1.
     
  13. May 20, 2016 #12
    Someone please explain it to me
     
  14. May 20, 2016 #13

    SammyS

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    Under what conditions is ##\ \lim_{x\to a} f(g(x)) = f(\lim_{x\to a} g(x)) \ ? ##
     
  15. May 20, 2016 #14
    I'm unaware about it please tell me
     
  16. May 20, 2016 #15

    SammyS

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    It's dangerous to use a theorem without paying attention to the conditions under which the theorem is valid.

    We can get back to that later. You might continue to ponder that.

    As an alternate way to understand this:
    Graph the function ##\displaystyle \ \left[ \frac{\sin(x-1)}{x-1} \right ] \ .##
     
  17. May 20, 2016 #16

    Mark44

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    Just to be clear, the function in question is ## \lfloor \frac{\sin(x-1)}{x-1}\rfloor##, the greatest integer function -- the largest integer less than or equal to ##\frac{\sin(x - 1)}{x - 1}##

    The LaTeX for this is ##\lfloor ##\dots## \rfloor##
     
  18. May 20, 2016 #17

    SammyS

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    Right. Since OP was using [ ] , I continued using that.

    With the standard floor symbols: ##\displaystyle \ \left\lfloor \frac{\sin(x-1)}{x-1} \right\rfloor \ ##
     
    Last edited: May 20, 2016
  19. May 20, 2016 #18

    Ray Vickson

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    Although the function ##f(h) = \sin(h)/h## has a limit of 1 as ##h \to 0## (where ##h = x-1##), the question is: how does ##f(h)## behave for ##h \neq 0## but near 0? Is it a little bit bigger than 1? Is it a little bit smaller than 1? You cannot answer these questions just from the limit alone; you need a bit more information.

    If ##f(h) < 1## for small ##|h| \neq 0## then ##1 < 2 f(h) < 2 ## for small, nonzero values of ##h##, and that will mean that ##\lfloor 2f(h) \rfloor = 1## for small nonzero values of ##h##.
     
  20. May 20, 2016 #19

    ehild

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    Evaluate ##\lfloor 2 \frac {\sin(x-1)}{x-1}\rfloor## for x=0.9, 0.99, 0.999, and x= 1.1, 1.01, and 1.001. What do you get? So what is the limit if x-->1?
    (Set your calculator to SCi digit 9)
     
    Last edited: May 21, 2016
  21. May 20, 2016 #20
    I don't know how to procede
     
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