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Homework Help: Find lim θ → 0: sinθ / θ + tanθ

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Find lim θ → 0: sinθ / θ + tanθ


    2. Relevant equations

    None.

    3. The attempt at a solution

    lim sinθ / θ + tanθ
    θ → 0

    = lim sinθ / θ + (sinθ / cosθ)
    θ → 0

    = lim sinθ / cosθ + sinθ / cosθ
    θ → 0

    = lim sinθcosθ / cosθ + sinθ
    θ → 0

    =sin(0)cos(0) / cos(0) + sin(0)

    = 0 * 1 / 1 + 0

    = 0

    The right answer is a 1/2. I dont know what I could have done wrong, as my method produces a reasonable answer of 0
     
  2. jcsd
  3. Apr 19, 2010 #2
    If you can't use TeX, please use parentheses to make things unambiguous. Are you trying to compute [tex]\lim_{\theta\to 0} \frac{\sin\theta}{\theta} + \tan\theta[/tex] or [tex]\lim_{\theta\to 0} \frac{\sin\theta}{\theta + \tan\theta}[/tex]?
     
  4. Apr 19, 2010 #3
    It should be the second one, ystael, because the limit of the first is 1.
     
  5. Apr 19, 2010 #4
    That's clear to me; I'm trying to get the OP to state his/her problem correctly.
     
  6. Apr 19, 2010 #5
    Its the second one. Sorry but I dont know how to use TeX. How can I use it so that I can not be ambiguous?
     
  7. Apr 19, 2010 #6
    Notation: If you need to write it on one line, you can write [tex]\lim_{\theta\to 0} (\sin\theta)/(\theta + \tan\theta)[/tex]; the top or bottom of a fraction must be parenthesized when it contains an operator that has a space in it (usually, anything except multiplications). Even [tex]\sin\theta / (\theta + \tan\theta)[/tex] (which is what I wrote in the first draft of this response) leaves some ambiguity: this could be taken to mean [tex]\sin \frac{\theta}{\theta + \tan\theta}[/tex].

    Your mistake above was in the transition from the second to the third line. In fact, because of the ambiguity in your fractions, I'm not sure what the third line is even supposed to mean, but that plain [tex]\theta[/tex] in the denominator just disappeared.

    Hint: Sometimes when you have a problem with an inconvenient fraction, you can make your life easier by computing with its reciprocal instead. Some argument may be necessary afterward to prove that taking 1 over your whole reasoning is justified.
     
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