lim V, m-->∞ =?

V = [(mg)/c][1-e^((-ct)/m)]

lim V m-->∞ =?

Please help me *understand* the basic method used to solve this. I tried ln on both sides and also a method with e, but fell short.

4 Answers

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  • 1 decade ago
    Best Answer

    lim V, m-->∞ =?

    V = [mg / c] [1 - e^(-ct)/m)]

    = g* [1 - e^(-ct)/m)] / [c / m] (0/0) form ------(1)

    let c / m = k

    V = g* [1 - e^-kt] / [k]

    --------------------------------------------------

    e^-kt = 1 - kt + (kt)^2 /2! - (kt)^3 / 3!+......

    [1 - e^-kt] = kt - (kt)^2 /2! + (kt)^3 / 3!+......

    divide both sides by k

    {[1 - e^-kt] / k } = t - k (t)^2 /2! + k^2(t)^3 / 3!+......

    ----------------------------------------------------

    put in (1)

    V = g*{t - k (t)^2 /2! + k^2(t)^3 / 3!+...} (use >> c / m = k)

    m-->∞ , k >>0 apply limit

    V = g*{ t - o (t)^2 /2! + o (t)^3 / 3!+...}

    V = g * t

    lim V, m-->∞ =?

    V = [(mg)/c][1-e^((-ct)/m)] = g * t answer

  • 1 decade ago

    V = (mg/c) [ ct/m +O(1/m^2) ] = gt + O(1/m) -->gt

    where O(m^r) means something/anything of the order of m^r

  • Anonymous
    1 decade ago

    This has indeterminate form ∞*0. Rewrite the expression as the second factor divided by c/(mg), which is the reciprocal of the first factor, and now you have indeterminate form 0/0. Use L'Hopital's rule on that.

  • Anonymous
    1 decade ago

    it's not easy !

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