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Can someone show me a straightforward proof that

$\lim_{m\to\infty}(1+\frac{r}{m})^{mt}=e^{rt}$

Thanks!

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1  
plug in $\tilde m = m/r$ and you're done –  yohBS Nov 25 '12 at 21:33
4  
@Dmitri what is your definition of $e^x$? –  user17762 Nov 25 '12 at 21:34
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2 Answers

up vote 3 down vote accepted

$$ \log\left(1+\frac rm\right)^{tm}=tm\,\log\left(1+\frac rm\right)=tm\,\left(\frac rm+O(\frac1{m^2})\right)=tr+O(\frac1m). $$ So $$ \left(1+\frac rm\right)^{tm}=e^{tr}\,e^{O(1/m)}. $$ Taking limit, the equality follows.

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I assume you meant to use "big-O", since your statement is false for "little-O". –  robjohn Nov 26 '12 at 10:21
    
Using $\log$ to answer this question is anachronistic. It's similar to using L'Hopital to show that $\lim\limits_{x\to0}\frac{\sin(x)}{x}=1$ to show that $\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)=\cos(x)$. –  robjohn Nov 26 '12 at 10:26
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In this answer, it is shown that $$ \lim_{m\to\infty}\left(1+\frac{r}{m}\right)^m=\lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{mr}\tag{1} $$ Since $x\mapsto x^t$ is a continuous function, raising $(1)$ to the $t$ power yields $$ \lim_{m\to\infty}\left(1+\frac{r}{m}\right)^{mt}=\lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{mrt}\tag{2} $$ Since $\lim\limits_{m\to\infty}\left(1+\frac1m\right)^m=e$ by definition, $(2)$ says $$ \lim_{m\to\infty}\left(1+\frac{r}{m}\right)^{mt}=e^{rt}\tag{3} $$

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