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Limit involving trigonometry

  1. Oct 17, 2014 #1
    1. The problem statement, all variables and given/known data
    "Calculate the following limit if it exists. If it does not exist, motivate why.
    [itex] \displaystyle\lim_{x\rightarrow 0} {\frac{x + x^2 +\sin(3x)}{tan(2x) + 3x}} [/itex]

    Do not use l'Hôpital's rule."



    2. Relevant equations
    [itex](1) \sin(a\pm b) = \cos(a)\sin(b)\pm\cos(b)\sin(a) [/itex]

    [itex](2) \cos(a\pm b) = \cos(a)\cos(b)\mp\sin(a)\sin(b) [/itex]

    [itex](3) \displaystyle\lim_{x\rightarrow 0} {\frac{\sin(x)}{x}} = 1 [/itex]

    [itex](4) \tan(x) = \frac{\sin(x)}{\cos(x)} [/itex]

    3. The attempt at a solution

    I have tried expressing the trigonometrics in terms of [itex]\sin(x)[/itex] and [itex]\cos(x)[/itex], but it just got messier without helping me in any way.

    This isn't me just jumping on these forums as soon as I can't find the answer; I have genuinely been trying to solve this problem and looking over my methods much more than once.

    Thanks in advance!
     
    Last edited: Oct 17, 2014
  2. jcsd
  3. Oct 17, 2014 #2

    LCKurtz

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    Try L'Hospital's rule.
     
  4. Oct 17, 2014 #3
    Ah! Forgot to mention: L'Hôpital's rule is prohibited on this exercise. Sorry, I'll add it to the post.
     
  5. Oct 17, 2014 #4

    ehild

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    Divide both the numerator and denominator by x. Use that the limit of sin(kx)/(kx) is zero if x goes to zero and k is a constant.

    ehild
     
  6. Oct 17, 2014 #5
    Thanks, I'll try that. Don't you mean the limit [itex] \displaystyle\lim_{x\rightarrow 0} {\frac{\sin(x)}{x}} = 1 [/itex] though?
     
  7. Oct 17, 2014 #6

    ehild

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    Yes, I meant that, but you have sin3x and sin(2x) so consider the limit of sin(kx)/x .
     
  8. Oct 18, 2014 #7
    Thank you for the help, I solved it yesterday by dividing numerator and denominator by 3x and realizing that 3x = 2x * 1.5.
     
  9. Oct 18, 2014 #8

    ehild

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    Clever! :)
     
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