Lets start with the most basic continuous interest formula and the evaluation of e:
Continuous interest: $\left(1+\frac{r}{n}\right)^n$
Evaluation of e: $\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x$
In this case, r is the interest rate (as a decimal) and n is the number of times or 'periods' in a year the invested amount will be compounded (for example, if the invested amount were compounded each day of the year, n would equal 365) Now lets substitute $x=\frac{n}{r}$. The formula now becomes:
$\left(1+\frac{1}{x}\right)^{xr}$
Based on the similarities between this formula and the evaluation of e, the formula can be rewritten as $e^r$, which is the first term in the most recognised variation of the formula for continuous interest, $e^r - 1$, but the question I have is: since we incorporate e into the formula for continuous interest, doesn't that mean we assume that the interest rate is always 100%? Since 100% = 1 (as a decimal), and using the evaluation of e, r is always equal to 1. Hence, why is it that this assumption is correct? Or is my logic flawed?
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$\begingroup$First, $(1+{1\over x})^x$ is not the same thing as $\lim_{x\rightarrow\infty}(1+{1\over x})^x$, so $e$ is not the base of your formula.
Second, continuous interest is not $(1+{r\over n})^n$; this is actually an improperly written formula for discrete-time compound interest factor. It should be correctly expressed as $(1+{r\over n})^{nt}$, where discrete time is $t$ and $n$ is compounding frequency. This expression can nevertheless be evaluated even if $t$ is continuous.
However, they're still related. The formula for proper continuous compound interest factor is $e^{rt}$. This is the consequence of
$$\lim_{n\rightarrow\infty}(1+{r\over n})^{nt}=e^{rt}$$
In words, as the compounding frequency becomes infinitely large such that the interest is compounded every infinitesimally small amount of time, then the Euler's constant actually comes into play. This feature is not by construction, but rather just an artifact of how we calculate interest, but it does make generalizing compound interest in continuous time easy.
To check if it's true, we can look at the instantaneous rate of return of $e^{rt}$:
$${d\over dt}\ln(e^{rt})={d\over dt}rt=r$$
It flips back to the interest rate, which is the rate of return.
$\endgroup$ 6 $\begingroup$Yes, the numerator in the limit for $e$ is $1$ rather than $r$ -- but note that the denominator is no longer $n$, so you have no particular reason to expect that the numerator in that fraction can be interpreted as (or ought to coincide with) a yearly interest rate.
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