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

Can someone show me a straightforward proof that

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

Thanks!

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

$$\log\left(1+\frac rm\right)^{tm}=tm\,\log\left(1+\frac rm\right)=tm\,\left(\frac rm+O(\frac1{m^2})\right)=tr+O(\frac1m).$$ So $$\left(1+\frac rm\right)^{tm}=e^{tr}\,e^{O(1/m)}.$$ Taking limit, the equality follows.
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
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 $$\lim_{m\to\infty}\left(1+\frac{r}{m}\right)^{mt}=e^{rt}\tag{3}$$