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Limit of a trig function

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data
    lim x->0 (2x+1-cosx)/(4x)


    2. Relevant equations



    3. The attempt at a solution

    factor out 1/4.
    get stuck because of cosx..
    and I'm not using l'hopital even though I know it.
     
  2. jcsd
  3. Oct 22, 2011 #2

    gb7nash

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    Homework Helper

    Have you learned Taylor series yet?
     
  4. Oct 22, 2011 #3

    Mark44

    Staff: Mentor

    I would bet that the OP hasn't learned Taylor series yet, but you don't need to use them, or L'Hopital's Rule, to evaluate this limit.

    Split the limit into two limits, one with (2x)/(4x) and the other with (1 - cosx)/(4x).

    There are a couple of special limits that are usually presented in textbooks in sections where there are limit problems involving trig functions. These are
    [tex]\lim_{x \to 0} \frac{sin(x)}{x} = 1[/tex]

    and
    [tex]\lim_{x \to 0} \frac{1 - cos(x)}{x} = 0[/tex]
     
  5. Oct 22, 2011 #4
    Thanks, I know L'hopital's but I knew there was a easier solution.

    By the way, what other special limits are there? Link would be nice!
     
  6. Oct 22, 2011 #5

    dynamicsolo

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    The two trigonometric limits that Mark44 provided are pretty much the "special" ones that people use. They crop up in developing the derivatives of sin x and cos x using "difference quotients". Those two are useful to know.

    The reason people generally don't bother looking for more of these is that once we do have L'Hopital's Rule, we calculate such trigonometric limits using that tool instead...
     
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