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Thread: Prove that lim...=e

  1. #1
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    Prove that lim...=e

    Prove that

    $\displaystyle \lim_{h \to 0}(1+h)^{1/h}=e$

    Hints outlined for this question are as follows.

    Since$\displaystyle \frac{d}{dx}\ln1=1$,
    As $\displaystyle h \to 0$, $\displaystyle \frac{\ln(1+h)-\ln1}{h}=\frac{\ln(1+h)}{h} \to 1$
    A second hint is that if $\displaystyle g$ is continuous at $\displaystyle c$ and $\displaystyle f$ is continuous at $\displaystyle g(c)$, then $\displaystyle f\circ g$ is continuous at $\displaystyle c$.

    You are not allowed to use L'Hospital's Rule to solve this question. This is one of the requirements, I'm afraid.
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  2. #2
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    Do you understand that $\displaystyle \lim _{n \to \infty } \left( {1 + \frac{1}{n}} \right)^n = e$?
    If so let $\displaystyle h=\frac{1}{n}$.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Do you understand that $\displaystyle \lim _{n \to \infty } \left( {1 + \frac{1}{n}} \right)^n = e$?
    If so let $\displaystyle h=\frac{1}{n}$.
    Didn't quite get what you were getting at at first, but then I realized it. Thanks.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Runty View Post
    Prove that

    $\displaystyle \lim_{h \to 0}(1+h)^{1/h}=e$

    Hints outlined for this question are as follows.

    Since$\displaystyle \frac{d}{dx}\ln1=1$,
    As $\displaystyle h \to 0$, $\displaystyle \frac{\ln(1+h)-\ln1}{h}=\frac{\ln(1+h)}{h} \to 1$
    A second hint is that if $\displaystyle g$ is continuous at $\displaystyle c$ and $\displaystyle f$ is continuous at $\displaystyle g(c)$, then $\displaystyle f\circ g$ is continuous at $\displaystyle c$.

    You are not allowed to use L'Hospital's Rule to solve this question. This is one of the requirements, I'm afraid.
    Let $\displaystyle 1=\lim_{x\to0}\frac{\ln(1+x)}{x}=\lim_{x\to0}\ln\l eft(\left(1+x\right)^{\frac{1}{x}}\right)=\ln\left (\lim_{x\to0}\left(1+x\right)^{\frac{1}{x}}\right)$. The last part follows from the hint stuff. The conclusion follows by exponentiation.
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