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Math Help - L'Hospitals Rule Troubles

  1. #1
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    L'Hospitals Rule Troubles

    Why does the limit as t approaches infinity of [(-te-5t)/5]-[e^-5t/25]= 1/25? I keep getting infinities when I try to use l'Hospitals rule.
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  2. #2
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    Quote Originally Posted by sfgiants13 View Post
    Why does the limit as t approaches infinity of [(-te-5t)/5]-[e^-5t/25]= 1/25? I keep getting infinities when I try to use l'Hospitals rule.
    Here is what you have posted:
    \lim _{t =  \to \infty } \left[ {\frac{{ - t}}<br />
{{5e^{5t} }} - \frac{1}<br />
{{25e^{5t} }}} \right] = \lim _{t =  \to \infty } \left[ {\frac{{ - 5t - 1}}<br />
{{25e^{5t} }}} \right] = \frac{{ - 1}}<br />
{{25}} Is that really what you meant?
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  3. #3
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    Quote Originally Posted by Plato View Post
    Here is what you have posted:
    \lim _{t =  \to \infty } \left[ {\frac{{ - t}}<br />
{{5e^{5t} }} - \frac{1}<br />
{{25e^{5t} }}} \right] = \lim _{t =  \to \infty } \left[ {\frac{{ - 5t - 1}}<br />
{{25e^{5t} }}} \right] = \frac{{ - 1}}<br />
{{25}} Is that really what you meant?


    I don't know how to use the math tags but the equatino above isn't right.

    (-1/5) x te^-5t - (1/25) x e^-5t as t approaches infinity.
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  4. #4
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    Quote Originally Posted by sfgiants13 View Post
    the equatino above isn't right.
    (-1/5) x te^-5t - (1/25) x e^-5t as t approaches infinity.
    Well it surely is the same to me.
    \left[ {\left( { - 1/5} \right)\left( {te^{ - 5t} } \right) = \frac{{ - t}}<br />
{{5e^t }}\,} \right]\& \,\left[ { - \left( {1/25} \right)\left( {e^{ - 5t} } \right) = \frac{{ - 1}}{{25e^{5t} }}} \right]
    What is incorrect about the above?
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  5. #5
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    I got -1/25 too for some reason but the book says it's convergant to 1/25. I'm thinking ti was just a math error in the book now...
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  6. #6
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    Quote Originally Posted by sfgiants13 View Post
    I got -1/25 too for some reason but the book says it's convergant to 1/25. I'm thinking ti was just a math error in the book now...
    \left[ {\left( { {\color{red}-} 1/5} \right)\left( {te^{ - 5t} } \right) = \frac{{ {\color{red}-} t}}{{5e^t }}\,} \right]
    I think that there is indeed a typo but I think it is in that first minus sign.
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