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Math Help - Prove A Limit.

  1. #1
    Junior Member qspeechc's Avatar
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    Prove A Limit.

    Prove that:

    (1 + x/n)^n = e^x as n approaches infinity

    Without using l'Hopitals Rule?

    The book proves (1 + 1/n)^n = e as n tends to infinity.
    by considering f(x) = ln(x) and f'(x) = 1/x and f'(1) = 1
    Then from first principles proves eventually that (1 + 1/n)^n = e as n tends to infinity, by letting n = 1/h etc.
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Start by giving a name to your limit, say

    \varphi  = \lim_{n \to \infty } \left( {1 + \frac{x}<br />
{n}} \right)^n.

    Since the logarithm is continuous on its domain, we can interchange the function and taking limits.

    \ln \varphi  = \lim_{n \to \infty } n\ln \left( {1 + \frac{x}<br />
{n}} \right).

    Substitute u=\dfrac1n,

    \ln \varphi  = \lim_{u \to 0} \frac{1}<br />
{u}\ln (1 + ux).

    Since \ln (1 + ux) = \int_1^{1 + ux} {\frac{1}<br />
{v}\,dv} ,\,1 \le v \le 1 + ux,

    we have \frac{1}<br />
{{1 + ux}} \le \frac{1}<br />
{v} \le 1\,\therefore \,\frac{x}<br />
{{1 + ux}} \le \frac{1}<br />
{u}\ln (1 + ux) \le x.

    Taking the limit as u\to0, we conclude by Squeeze Theorem that \lim_{u \to 0}\frac{1}<br />
{u}\ln (1 + ux) = x, and we happily get

    \ln \varphi  = x\,\therefore \,\varphi  = e^x .
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  3. #3
    Global Moderator

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    Quote Originally Posted by Krizalid View Post
    Start by giving a name to your limit, say

    \varphi  = \lim_{n \to \infty } \left( {1 + \frac{x}<br />
{n}} \right)^n.
    No! That is not good. You are assuming the limit exists. How do you know the limit exists?

    What you should do is write,
    \left( 1 + \frac{x}{n} \right)^n = \exp \left[  n\ln \left( 1 + \frac{x}{n} \right) \right]

    Since \ln (1 + ux) = \int_1^{1 + ux} {\frac{1}<br />
{v}\,dv} ,\,1 \le v \le 1 + ux,

    we have \frac{1}<br />
{{1 + ux}} \le \frac{1}<br />
{v} \le 1\,\therefore \,\frac{x}<br />
{{1 + ux}} \le \frac{1}<br />
{u}\ln (1 + ux) \le x.
    Here you are assuming that x>0. It does not cover the case when x<0.
    (The way I have it above also does x>0).

    Anyway, good job.
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