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L'hospitals rule

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data
    evaluate the limit
    [tex] (lim.t\rightarrow0) \frac{e^6^t-1}{t} [/tex]


    2. Relevant equations
    l'hospital's rule i guess


    3. The attempt at a solution
    With the usual approach of this rule, you are suppose to take a derivative and evaluate the derivative at that original limit. The problem is that every time I take a derivative and apply the limit I keep getting an indeterminate form. I don't know how to solve this. Can someone point me in the right direction?

    Thanks
     
  2. jcsd
  3. Nov 20, 2009 #2
    Once you use l'Hôpital's rule once, it's not in an indeterminate form and you can find the limit. Try showing your work.
     
  4. Nov 20, 2009 #3
    ok so

    [tex] \frac {d}{dx} (\frac {e^6^t-1}{t})

    = \frac{t(6e^6^t)-(e^6^t-1)}{t^2}

    = \frac {6te^6^t-e^6^t+1}{t^2}[/tex]

    apply the limit

    [tex] \frac{0-1+1}{0}= \frac {0}{0} [/tex]
     
  5. Nov 20, 2009 #4
    When using l'Hôpital's rule, you take the derivative of the top and bottom separately, do not use the difference [STRIKE]quotient[/STRIKE] rule!
    [tex]\lim_{x\rightarrow 0}\frac{f(x)}{g(x)} = \lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)}[/tex]
     
    Last edited: Nov 20, 2009
  6. Nov 20, 2009 #5
    oh so NO quotient rule? oh ok. let me redo it then hold on.
     
  7. Nov 20, 2009 #6
    well just looking at it the answer is 6. wow thanks for that. I thought you had to use the quotient rule.

    many thanks
     
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