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Question

# Suppose that P dollars is invested at an annual interest rate of$r \times 100 \%$. If the accumulated interest is credited to the account at the end of the year, then the interest is said to be compounded annually; if it is credited at the end of each 6-month period, then it is said to be compounded semiannually; and if it is credited at the end of each 3-month period, then it is said to be compounded quarterly. The more frequently the interest is compounded, the better it is for the investor since more of the interest is itself earning interest. One can imagine interest to be compounded each day, each hour, each minute, and so forth. Carried to the each hour, each minute, and so forth. Carried to the instant of time; this is called continuous compounding. Thus the value A of P dollars after t years when invested at an annual rate of$r \times 100 \%$, compounded continuously, is$A = \lim _ { n \rightarrow + \infty } P \left( 1 + \frac { r } { n } \right) ^ { n t }$. Use the fact that$\lim _ { x \rightarrow 0 } ( 1 + x ) ^ { 1 / x } = e$to prove that$A = P e ^ { r t }$.

Solution

Verified
Step 1
1 of 3

Given information and knowing what we want

$\begin{equation*} A=\lim_{n \rightarrow +\infty} P \Big(1+\frac{r}{n}\Big)^{nt} \end{equation*}$

And

$\begin{equation} \lim_{x \rightarrow 0} (1+x)^{1/x} =e \end{equation}$

Claim 1:

$\begin{equation*}A=Pe^{rt}\end{equation*}$

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