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anonymous

  • 5 years ago

CALCULUS QUESTION: Using l'Hopital's Rule evaluate the following limit: lim (1 + 5/x)^(x/4) as x ->+infinity

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  1. anonymous
    • 5 years ago
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    when you substitute in infinity for x you first get 1^(infinity) which is indeterminate. So you have to manipulate the equation. If you set the limit equal to L and take the ln of both sides you end up with ln(L) = lim x-> infinity [(1+5/x)]^(x/4) which is equivalent to ln(L) = lim x-> infinity (x/4)[(1+5/x)]

  2. anonymous
    • 5 years ago
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    This is now at the stage of an infinity times 0. So put the x in the denominator to get ln(L) = lim x-> infinity (1/4)ln[(1+5/x)]/x^(-1) to force a 0/0 case and take the derivatives of the numerator and denominator to get a messy equation to simplifies to ln(L) = lim x-> infinity 5/[4(1 + 5/x)] or just ln(L) = 5/4 so L = e^(5/4)

  3. anonymous
    • 5 years ago
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    If you remember that lim x->infinity (1 + 1/x)^(x) = e then the solution should make sense even if the explanation is confusing.

  4. anonymous
    • 5 years ago
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    yeah that makes sense, thank you

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