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Math Help - L'Hopitals Rule

  1. #1
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    L'Hopitals Rule

    here's the problem:

    Find the limit of lim (e^t)-1 / t^3
    t-->0

    I thought to use L'Hopital's Rule 3 times to get lim e^t / 6
    t-->0

    Then it would no longer be indeterminate (0/0) and I would have 1/6 as my answer.

    But my book says the answer is infinity.

    What did I do wrong?
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  2. #2
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    Quote Originally Posted by mech.engineer.major View Post
    here's the problem:

    Find the limit of lim (e^t)-1 / t^3
    t-->0

    I thought to use L'Hopital's Rule 3 times to get lim e^t / 6
    t-->0

    Then it would no longer be indeterminate (0/0) and I would have 1/6 as my answer.

    But my book says the answer is infinity.

    What did I do wrong?

    You need to check the hypothesis each time you use L.H's rule.

    After one application you get

    \lim_{t \to 0}\frac{e^{t}-1}{t^3}=\lim_{t \to 0}\frac{e^{t}}{3t^2}

    This limit is not of the form \frac{0}{0} or \frac{\infty}{\infty} so we CANNOT use L.H's rule.


    Now if we take the limit the numerator goes to 1 and the denominator
    goes to zero from both sides so

    \lim_{t \to 0}\frac{e^{t}-1}{t^3}=\lim_{t \to 0}\frac{e^{t}}{3t^2}=\infty
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  3. #3
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    wait I thought dividing by zero doesn't exist. you're saying 1/0 is infinity?
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  4. #4
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    Quote Originally Posted by mech.engineer.major View Post
    wait I thought dividing by zero doesn't exist. you're saying 1/0 is infinity?
    You can't divide by 0, but you can see what happens when a function has a denominator that TENDS to 0.
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  5. #5
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    okay I graphed it on my calculator so now it makes sense. is it always a good idea to check like that?
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  6. #6
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    Quote Originally Posted by mech.engineer.major View Post
    okay I graphed it on my calculator so now it makes sense. is it always a good idea to check like that?
    Absolutely.
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  7. #7
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    thank you both for your help
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