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Limit of a Sin Function

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP120619i0ea2h15bci31300005d9a3422292h43h4?MSPStoreType=image/gif&s=43&w=174&h=39 [Broken]

    2. Relevant equations



    3. The attempt at a solution

    So Wolfram says to use L'Hopital as the first step, we haven't learned anything about this yet so there has to be another way using calculus that a first year math student would know.

    I would first expand the denominator of x^2-4 to (x+2)(x-2) and cancel out the x+2 in both numerator and denominator. And I'm left with:
    ((x-1)sin1)/(x-2)
    This is where I get lost...

    Any help would be much appreciated!
    Jim
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 5, 2011 #2
    One cannot seperate (x+2) from sin any more than one can seperate ... a bone from the mouth of a very hungry dog!

    Either one 'cancels' the whole sin(x+2) or nothing at all.
     
  4. Nov 5, 2011 #3

    LCKurtz

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    You don't cancel the (x+2)'s. You should know

    [tex]\lim_{x\rightarrow 0}\frac{\sin x}{x}[/tex]

    Use that somehow.
     
  5. Nov 5, 2011 #4
    Did you mention in class that lim(a x b) = lim(a) x lim(b)?
     
  6. Nov 5, 2011 #5
    Yes, we mentioned all the above, and this was much easier after LCKurtz's suggestion:

    =lim sin((x+2)/(x+2)) * lim (x-1)/(x-2)
    = 1 * -3/-4
    =3/4
    I really need to learn how to use that equation generator thing...

    Much appreciated folks! :)
     
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