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Thread: 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

    $\displaystyle \varphi = \lim_{n \to \infty } \left( {1 + \frac{x}
    {n}} \right)^n.$

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

    $\displaystyle \ln \varphi = \lim_{n \to \infty } n\ln \left( {1 + \frac{x}
    {n}} \right).$

    Substitute $\displaystyle u=\dfrac1n,$

    $\displaystyle \ln \varphi = \lim_{u \to 0} \frac{1}
    {u}\ln (1 + ux).$

    Since $\displaystyle \ln (1 + ux) = \int_1^{1 + ux} {\frac{1}
    {v}\,dv} ,\,1 \le v \le 1 + ux,$

    we have $\displaystyle \frac{1}
    {{1 + ux}} \le \frac{1}
    {v} \le 1\,\therefore \,\frac{x}
    {{1 + ux}} \le \frac{1}
    {u}\ln (1 + ux) \le x.$

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

    $\displaystyle \ln \varphi = x\,\therefore \,\varphi = e^x .$
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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

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

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

    Since $\displaystyle \ln (1 + ux) = \int_1^{1 + ux} {\frac{1}
    {v}\,dv} ,\,1 \le v \le 1 + ux,$

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

    Anyway, good job.
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