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Solving trigonometric limits

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    solve the following limit: [tex]\lim_{x\rightarrow0}\frac{x+sinx}{2x-sin3x}[/tex]

    3. The attempt at a solution

    I know the principle of solving all sorts of limits but not trig limits, and since we just started trig limits it's still not clear to me, but i think to solve it we have to separate it then solve but i don't know. Any help on how to start please?
  2. jcsd
  3. Feb 21, 2012 #2


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    Do you know l'Hopital's rule?
  4. Feb 21, 2012 #3
    No haven't learned it can you fill me on it please?
  5. Feb 21, 2012 #4


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    Are you familiar with Taylor series ?

    Added in Edit:

    Better yet:
    multiply by (1/x) over (1/x) .​
  6. Feb 21, 2012 #5


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    You may not have covered it yet so you may be expected to do it a different way. Divide the numerator and denominator by x. Now you probably do know the limit sin(x)/x. Try to use that to find the limits of the terms.
  7. Feb 21, 2012 #6
    No sorry i'm not familiar with taylor series, but i know that for l'hospital's rule we have to take the derivative of the numerator and the derivative of the denominator; but i have never learned derivatives so i can't do that. Is it possible to solve it some other way?
  8. Feb 21, 2012 #7
    Ok the derivative of x+sinx is cosx+1 and how about the bottom?
  9. Feb 21, 2012 #8
    ok i counted and found the limit to be -2 is that correct?
  10. Feb 21, 2012 #9


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    Yes, if you used l'Hopital then you might check whether that is allowed. Otherwise you should try doing it using that the limit of sin(x)/x is 1.
  11. Feb 21, 2012 #10
    Yea thank you very much I get the trick. It's the same for almost all of them.
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