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Homework Help: How to do this limit?

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - sinx}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Do I need to do a l'hopital?
     
  2. jcsd
  3. Mar 23, 2010 #2

    Dick

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    Why are you asking? It's 0/0 form, isn't it?
     
  4. Mar 24, 2010 #3
    What do I do with a [tex]e^{-0}[/tex]? Is that the same as just 0
     
  5. Mar 24, 2010 #4

    Dick

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    Sure. -0 is the same as 0. What's e^0?
     
  6. Mar 24, 2010 #5
    1
    Ok so I do some l'hops and I get this

    [tex]\frac {e^x + e^{-x} - x}{1 - cosx} = \frac {e^x - e^{-x} - 1}{sinx} = \frac {1-1-1}{1} = -1[/tex]
     
  7. Mar 24, 2010 #6

    Dick

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    Look at your first step. Since when is the derivative of 2x equal to x?
     
  8. Mar 24, 2010 #7

    Mark44

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    Just to clarify, e-0 = e0 = 1. I believe Dick understood what you meant, zeion, even though you didn't ask the question clearly. IOW e-0 is not the same as 0.
     
  9. Mar 24, 2010 #8
    Ok sorry so I get like
    [tex]\frac {e^x + e^{-x} -2}{1-cosx} = \frac {e^x - e^{-x}}{sinx} = \frac {e^x + e^{-x}}{cosx} = 2?[/tex]
     
  10. Mar 24, 2010 #9
    I feel like I'm reading a book called "Mathematical Calculations in the Twilight Zone". :smile:
     
  11. Mar 24, 2010 #10

    Dick

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    That's the right answer, but elect_eng makes a good point. Those quantities AREN'T equal. The LIMITS as x->0 of those quantities are equal. That's what makes it spooky to read.
     
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