How is $ \lim_{t \to \infty}\left(1+\frac{r}{t}\right)^{tn} = e^{rn} $. How is this limit computed? $r$ is a constant.
$$ \lim_{t \to \infty}\left(1+\frac{r}{t}\right)^{tn}$$
 A: Change the variable in the limit by $x=\frac tr$. We know that
$$
\lim_{x \to \infty}\left(1+\frac{1}{x}\right)^{x} = e.
$$
Now if $t\to\infty$, we can say that $x\to\infty$ and we have:
$$
\lim_{t \to \infty}\left(1+\frac{r}{t}\right)^{tn}= \lim_{x \to \infty}\left(1+\frac{1}{x}\right)^{xrn} = \lim_{x \to \infty}\left(\left(1+\frac{1}{x}\right)^{x}\right)^{rn}  =e^{rn}
$$
A: A very basic limit is
$$
\lim_{t\to\infty}\left(1+\frac{r}{t}\right)^t=e^r
$$
Since raising to the $n$-th power is continuous, you have
$$
\lim_{t\to\infty}\left(1+\frac{r}{t}\right)^{tn}=
\lim_{t\to\infty}\left(\left(1+\frac{r}{t}\right)^t\right)^n=
\left(\lim_{t\to\infty}\left(1+\frac{r}{t}\right)^t\right)^n=(e^r)^n=
e^{rn}
$$
A: Expand the term $(1 + \frac{r}{t})^{tn}$ by binomial theorem for finite $t$ and then take $t \rightarrow \infty$.
So $$(1 + \frac{r}{t})^{tn} = 1 + {tn \choose 1} \frac{r}{t} + {tn \choose 2} (\frac{r}{t})^2 + \dots \\ = 1 + \frac{tn}{1!}\frac{r}{t} + \frac{tn(tn - 1)}{2!}(\frac{r}{t})^2 + \dots \\ = 1+ \frac{rn}{1!} + \frac{rn(rn - \frac{1}{t})}{2!} + \dots$$
For finite $t$ you shall get finite terms. Now take $t \rightarrow \infty $ and find out the limit.
A: Call the limit y. Take logarithms from both sides. On the LHS use L Hopital's rule. now calculate y by using the definition of a logarithm.
