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Limit - Squeeze Theorem or L'Hopital's Rule?

  1. Oct 7, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data



    2. Relevant equations

    Squeeze theorem: set up inequalities putting the function of interest between two integers.

    L'Hopital's rule: when plugging in the number into the limit results in a specified indeterminate form such as 0/0 or infinity/infinity then take the derivative of the numerator and the derivative of the denominator and again plug-in to solve the limit.

    The limit of 1-cosx/x as x --> 0 is 1.

    3. The attempt at a solution

    I used L'Hopital's rule and applied two iterations of get

    cos(x)/(4cosx+4cosx-4xsinx)

    Plugging in 0 results in

    1 / (4+4-0) = 1/8

    However, we never learned L'Hopital's rule in class and the only thing we learned was the squeeze theorem. I attempted to use the squeeze theorem (without success):

    0≤1-cosx≤2

    I'm not sure where to proceed beyond the above step in using the squeeze theorem.

    0/x≤(1-cosx)/x≤2/x
    x--> 0 so infinity≤(1-cosx)/x≤infinity

    Is the squeeze theorem the correct method to solve this problem? Or am I overlooking some common identity or method other than L'Hopital's rule to solve this problem?
     
  2. jcsd
  3. Oct 7, 2013 #2

    LCKurtz

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    Not generally between two integers, just two numbers.

    I assume you mean ##\frac{1-\cos x}{x}## and not what you have written:##1 -\frac{\cos x}{x}##. Use parentheses! Also, you will have difficulty proving it since the limit isn't ##1##.

    How in the world did you get that with L'Hospital's rule?? Show your work.

    ##\frac 0 x = 0## so you have ##0## on the left and infinity on the right. Nothing is being squeezed. Try doing L'Hospital's rule correctly.
     
  4. Oct 7, 2013 #3
    If you use L'Hospital's rule you calculate the derivative of 1-cosx and x, which gives you sinx/1 or simply sinx. What's the limit as x approaches 0 of sinx?
    Another method to evaluate this limit is using the fact that the limit as x approaches 0 of sinx/x is 1, and then multiplying your original problem ((1-cosx)/x) by the conjugate of the numerator. When you simplify this you'll see the limit I mentioned earlier pop up which helps you simplify and get to the correct answer.
     
  5. Oct 7, 2013 #4
    I think you took the derivative of the whole functions, you're supposed to take the derivative of the top and bottom seperately. So the derivative of the top/derivative of the bottom. The bottom gives you 1 and the top gives you what?
     
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