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Problem 79 Hard Difficulty

If an initial amount $ A_0 $ of money is invested at an interest rate $ r $ compounded $ n $ times a year, the value of the investment after $ t $ years is
$$ A = A_0 \left( 1 + \frac{r}{n} \right)^{nt} $$
If we let $ n \to \infty $, we refer to the continuous compounding of interest. Use l'Hospital's Rule to show that if interest is compounded continuously, then the amount after $ t $ years is
$$ A = A_oe^{rt} $$

Answer

First we will find $\lim _{n \rightarrow \infty}\left(1+\frac{r}{n}\right)^{n t},$ which is of the form $1^{\infty} \cdot y=\left(1+\frac{r}{n}\right)^{n t} \Rightarrow \ln y=n t \ln \left(1+\frac{r}{n}\right),$ so
\[
\begin{array}{l}
\lim _{n \rightarrow \infty} \ln y=\lim _{n \rightarrow \infty} n t \ln \left(1+\frac{r}{n}\right)=t \lim _{n \rightarrow \infty} \frac{\ln (1+r / n)}{1 / n} \stackrel{n}{=} t \lim _{n \rightarrow \infty} \frac{\left(-r / n^{2}\right)}{(1+r / n)\left(-1 / n^{2}\right)}=t \lim _{n \rightarrow \infty} \frac{r}{1+i / n}=t r \Rightarrow \\
\lim _{n \rightarrow \infty} y=e^{r t} . \text { Thus, as } n \rightarrow \infty, A=A_{0}\left(1+\frac{r}{n}\right)^{n t} \rightarrow A_{0} e^{r t}
\end{array}
\]

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Video Transcript

So here we have um an interest problem. So A equals A. Not Times one Plus R. Divided by and to the empty. Um So is constant. So we don't really care about it in terms of the limit, we can remove it when we take the limit as N. Goes to infinity. So based on this we can take the natural log of both sides to simplify things. When we rewrite it, we end up getting the natural log of one plus R over N divided by one over Mt. Then when we plug in and going to infinity we need to use towels real again. So we end up getting R. P. by one plus are over 10. So now when we plug an ongoing to infinity we get artie over one. So it's just going to be our t remember we took the natural log of both sides of the natural log of Y equals RT. Which means that why is going to equal E. To the R key? However, we did remove that A not So we have to put that back into the original equation, meaning that A. Is going to equal a. Not. E. To the RT.

Carson M.

California Baptist University

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Problem 79 Hard Difficulty

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