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Limit of (1 - cosh(2x)) / 4x^3 + x^2

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\mathop {\lim }\limits_{x \to 0 } \frac{1 - cosh(2x)}{{4x^3 + x^2}}[/tex]


    2. Relevant equations

    Product, sum, quotient laws


    3. The attempt at a solution

    [tex]\mathop {\lim }\limits_{x \to 0 } \frac{1 - cosh(2x)}{{4x^3 + x^2}} =
    \mathop {\lim }\limits_{x \to 0 } \frac{\lim 1 - \lim cosh(2x)}{{\lim 4 + \lim x^3 + \lim x^2}}
    =
    \mathop {\lim }\limits_{x \to 0 } \frac{1 - \lim cosh(2x)}{{4 + 0 + 0}}
    =
    \mathop {\lim }\limits_{x \to 0 } \frac{1 - cosh(\lim 2x)}{{4}}
    =
    \frac{1 - cosh(0)}{{4}}
    = \frac{1 - 1}{{4}}
    = 0
    [/tex]

    However the answer is supposed to be 2.

    I was sure the denominator should be 4, but not sure how to get 8 in the numerator :confused:
     
  2. jcsd
  3. May 25, 2009 #2

    Gib Z

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    Homework Helper

    Okay well, You seem to have changed that 4x^3 to 4 + x^3, which is how you got your zero denominator to be 4.

    Firstly, do you know [tex] \lim_{x\to 0} \frac{ \sinh x}{x} [/tex] ? That is a good thing to know and is easy to find.

    Also, can you express [tex]\cosh 2x[/tex] in terms of sinh^2 x ? That will help.
     
  4. May 25, 2009 #3
    sinh(0) = 0 so use LHopitals rule :
    [tex] \lim_{x\to 0} \frac{ \sinh x}{x} = \lim_{x\to 0} \frac{ \cosh x}{1} = \frac{cosh 0}{1} = \frac{1}{1} = 1 [/tex]

    I don't know what you mean, sorry.
     
  5. May 25, 2009 #4

    Gib Z

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    Homework Helper

    Ok well, have you seen the identity

    [tex]\cosh 2x = 2\sinh^2 x +1[/tex]
    ?
     
  6. May 25, 2009 #5
    I tried subbing x = 0 into the above in the original equation and get:

    [tex]\frac{1 - 2e^2 + 4 - \frac{2}{e^2}}{16x^3 + 4x^2}[/tex]

    Now should I divide each term by [tex]x^3[/tex]?

    Can someone please show me the method? I am getting confused now.
     
  7. May 25, 2009 #6

    Gib Z

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    Homework Helper

    Directly substitute that identity into your limit, and apply the other limit you worked out.
     
  8. May 25, 2009 #7
    You mean into this? [tex]\mathop {\lim }\limits_{x \to 0 } \frac{\lim 1 - \lim cosh(2x)}{{\lim 4 + \lim x^3 + \lim x^2}}[/tex]

    so, [tex]\mathop {\lim }\limits_{x \to 0 } \frac{\lim 1 - \lim 2 + \lim sinh^2(x+1)}{{\lim 4 + \lim x^3 + \lim x^2}} = \frac{1 - 2 + \lim sinh^2(x+1)}{4} [/tex]

    I can't get that to equal 2.
     
  9. May 25, 2009 #8
    Hey dude apply l-hospital rule when there is 0/0 form after putting x=0 in the original equation , now differentiate the equation not in u/v form differentiate upper equation and lower equation differently until u got the equation not in the form 0/0 after putting the value
     
  10. May 25, 2009 #9
    First use L'Hopital's rule with the original problem and keep using it until you're done. You probably want to use an identity to make differentiation a little easier after one step.
     
  11. May 25, 2009 #10
    Best answer I can get is 1/2 or -1/2.

    Anyway test on this starts in 45 minutes so just hope this question doesn't come up.
     
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