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Limit of trigonometric function

  1. Aug 11, 2014 #1
    1. The problem statement, all variables and given/known data

    lim x --> 0 for function y = (-2x)/(sinx)

    Using L'Hopital's Theorem, I found the derivative of the top and of the bottom and found the limit and got -2. How to find the limit as x approaches 0 without using L'Hopital's theorem. I know my solutions manual uses a different method, but I don't know what it is because they have skipped all those steps.
     
  2. jcsd
  3. Aug 11, 2014 #2
    If they skipped all the steps, how do you know that they did it differently from you? Is it possible that they expanded the denominator in a Taylor series about x = 0?

    Chet
     
  4. Aug 11, 2014 #3

    jbunniii

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    No matter how you solve this problem, you will effectively need to establish that
    $$\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$$
    There are many ways to do this, L'Hospital's rule being the simplest but not the most elementary.

    If you know about Taylor series, then you can expand ##\sin(x)##, divide each term by ##x##, and take the limit. All terms except the first will go to zero.

    There's also nice proof which uses a geometric (not 100% rigorous, but very intuitively convincing) argument to establish the following inequality, which is valid for all nonzero ##x \in [-\pi/2, \pi/2]##:
    $$\cos(x) \leq \frac{\sin(x)}{x} \leq 1$$
    The result then follows from the squeeze theorem. See the first answer at this StackExchange link to see the picture from which the inequality is inferred:

    http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1
     
  5. Aug 11, 2014 #4
    The simplest way (the first week of Calc 1) to solve this problem is to note that (-2x)/sin(x) =-2/(sin(x)/x) then solve the problem using standard limit rules.
     
  6. Aug 14, 2014 #5
    Thank you all for posting. I ended up using shortbus_bully's method because he was right it was the simplest way to solve the problem!
     
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