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

Solution 1:

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

Solution 2:

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 $$ \lim_{m\to\infty}\left(1+\frac{r}{m}\right)^{mt}=e^{rt}\tag{3} $$