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Math Help - Limits using L'Hospital's rule.

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    Limits using L'Hospital's rule.

    Evaluate the limit using L'Hospital's rule.

    y = lim x->infinity (5x-1/5x+3)^5x+2


    ln y = (5x + 2) ln (5x-1/5x+3)

    lim x-> infinity lny= lim x-> infinity ln (5x-1)-ln (5x+3)/ 1/(5x+2)= lim x-> infinity 5/(5x − 1) − 5/(5x + 3)/−5/(5x + 2)^2

    I'm not sure how to proceed after this step.

    Any help would be appreciated, thank you for your time and effort
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    Quote Originally Posted by ioke09 View Post
    Evaluate the limit using L'Hospital's rule.

    y = lim x->infinity (5x-1/5x+3)^5x+2


    ln y = (5x + 2) ln (5x-1/5x+3)

    lim x-> infinity lny= lim x-> infinity ln (5x-1)-ln (5x+3)/ 1/(5x+2)= lim x-> infinity 5/(5x − 1) − 5/(5x + 3)/−5/(5x + 2)^2

    I'm not sure how to proceed after this step.

    Any help would be appreciated, thank you for your time and effort
    Note that \displaystyle \frac{5x - 1}{5x + 3} = 1 - \frac{4}{5x + 3}. Therefore l'Hospital's Rule is not required if you're familiar with the basic limit \displaystyle \lim_{t \to +\infty} \left(1 + \frac{a}{t} \right)^t = e^a.

    In your limit, substitute t = 5x + 3, re-arrange using a basic index law and then use the limit of a product theorem. It becomes:

    \displaystyle \lim_{t \to +\infty} \left[\left(1 - \frac{4}{t} \right)^{-1} \left(1 - \frac{4}{t} \right)^t\right] = e^{-4}.
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