How is $ \lim_{t \to \infty}\left(1+\frac{r}{t}\right)^{tn} = e^{rn} $.

Question

How is this limit computed? $r$ is a constant. $$ \lim_{t \to \infty}\left(1+\frac{r}{t}\right)^{tn}$$

Answer 1

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: $$ \lim_{t \to \infty}\left(1+\frac{r}{t}\right)^{tn}= \lim_{x \to \infty}\left(1+\frac{1}{x}\right)^{xrn} = \lim_{x \to \infty}\left(\left(1+\frac{1}{x}\right)^{x}\right)^{rn} =e^{rn} $$

Answer 2

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 $$ \lim_{t\to\infty}\left(1+\frac{r}{t}\right)^{tn}= \lim_{t\to\infty}\left(\left(1+\frac{r}{t}\right)^t\right)^n= \left(\lim_{t\to\infty}\left(1+\frac{r}{t}\right)^t\right)^n=(e^r)^n= e^{rn} $$

Answer 3

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.

Answer 4

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.