Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

what does (1+r/m)^m tend to as m tends to infinity? Thanks!?

Update:

I think the last answer is right but can anyone explain why

(1+1/(m/r))^r(m/r)

is e^r?

Thanks!

3 Answers

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  • Anonymous
    1 decade ago
    Best Answer

    (1+r/m)^m

    (1+1/(m/r))^m

    (1+1/(m/r))^mr/r

    (1+1/(m/r))^r(m/r)

    as m -> infin, so does m/r

    (1+1/(m/r))^r(m/r)

    This is e^r

    Lets evaluate the limit mathematically.

    lim (1+r/m)^m

    e^ln lim (1+r/m)^m

    e^lim ln (1+r/m)^m

    e^lim m ln(1+r/m)

    e^lim ln(1+r/m) / (1/m)

    e^lim d/dm ln(1+r/m) / d/dm (1/m)

    e^lim r d/dm(1/m) / [(1+r/m) d/dm (1/m) ]

    e^lim r / (1+r/m)

    e^r

    =====

    Hello.

    The last answer is right? No... my answer is right. In fact, my answer is the EXACT SAME answer as the last person. The last answerer repeated exactly what I stated. So dont you dare give that person credit for what I already told you.

    The limit as h → ∞

    lim_{h → ∞} (1 + 1/h)^h = e

    This is the definition of 'e'

    And the above proof in the first half of this post explains why your limit is e^r

  • 1 decade ago

    1. As m tends to infinity, the term r/m tends to 0 since m becomes larger and larger. Hence the term r/m gets smaller and smaller until it tends to 0.

    So (1+ r/m) tends to 1 when m tends to inifinity. And we all know that 1 raised to the power of any value is still 1.

  • Anonymous
    1 decade ago

    As m tends to infinity, r/m tends to 0 and so the bit in the brackets tends to1.

    As m tends to infinity therefore the equation tends to 1^m = 1

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