# Limits at infinity for compound interest

I have been revising for my notes and encountered this formula:

\begin{align*} \lim_{m\rightarrow \infty}A\begin{pmatrix}1+\frac r m\end{pmatrix}^{mt} &=\lim_{m\rightarrow \infty}A\begin{pmatrix}1+\frac 1 {m/r}\end{pmatrix}^{(m/r)rt}\\ &=A\begin{bmatrix}\lim_{m\rightarrow \infty}\begin{pmatrix}1+\frac 1 {m/r}\end{pmatrix}^{(m/r)}\end{bmatrix}^{rt}\\ &=Ae^{rt} \end{align*}

The puzzling part is how do I transit on the first line? In other words, how did this happen? \begin{align*} \lim_{m\rightarrow \infty}A\begin{pmatrix}1+\frac r m\end{pmatrix}^{mt} &=\lim_{m\rightarrow \infty}A\begin{pmatrix}1+\frac 1 {m/r}\end{pmatrix}^{(m/r)rt}\\ \end{align*} I have been trying to divide by $m/r$ on both sides of the equation but do not know how the power comes in. Any ideas?

UPDATE: solved

\begin{align*} \lim_{m\rightarrow \infty}A\begin{pmatrix}1+\frac r m\end{pmatrix}^{mt} &=\overbrace{\lim_{m\rightarrow \infty}A\begin{pmatrix}1+\frac 1 {m/r}\end{pmatrix}^{mt}}^{ \text{Divide by }\space r \space \text{on both the numerator & denominator}}\\ &=\overbrace{\lim_{m\rightarrow \infty}A\begin{pmatrix}1+\frac 1 {m/r}\end{pmatrix}^{(m/r)rt}}^{ \text{Divide & then multily by }\space r \space \text{on the power}}\\ \end{align*}

• Just note that $\frac{r}{m} = \dfrac{1}{\frac{m}{r}}$ (dividing by a fraction is multiplying by its reciprocal), and $m = \frac{m}{r} \cdot r$. – JavaMan Feb 28 '13 at 3:57

## 1 Answer

Two typos. The author(s) did not mean to write $\frac{r}{m/r}$, what was intended is $\frac{1}{m/r}$. And $(m/r)rt$ was intended, not $(m/r)mt$.

Edit: Typo has now been corrected. There is no transition in the first line. The fraction $\frac{1}{m/r}$ is exactly the same as $\frac{r}{m}$. Similarly, $(m/r)rt$ is exactly the same thing as $mt$.

The expression is being rewritten as an equivalent expression so that we can use the fact that $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$.

• Hello, thank you for helping me spot the typos! I have made the necessary changes. However, I still can't understand the transition part. Could you help? – bryanblackbee Feb 28 '13 at 3:55
• I feel very cheated right now...finally got it, thank you! – bryanblackbee Feb 28 '13 at 4:02
• You are welcome. Basically, in the first line, nothing was happening, but it was happening in a useful way. – André Nicolas Feb 28 '13 at 4:05
• Yes, I realized! Updated the solution to address that. – bryanblackbee Feb 28 '13 at 4:07