From Periodic Compounding to Continuous Growth
A balance compounded at regular intervals changes in steps: interest is added at the end of each compounding period. If the number of periods per year increases, each period becomes shorter and the interest added at each step becomes smaller. The limit of this process leads to a smooth model of continuous growth.
Let \(P_0>0\) be an initial amount, let \(r\) be a fixed annual rate written as a decimal, and let \(t\geq 0\) be time in years. For example, an annual rate of \(6\%\) means \(r=0.06\). If interest is compounded \(n\) times per year, the rate per compounding period is \(r/n\).
The distinction in the definition matters. In an actual discrete schedule, the number of completed periods must be an integer. But the expression with a real exponent is also mathematically defined when \(1+r/n>0\), whether or not a schedule has exactly \(nt\) periods at time \(t\). The limit below is first an algebraic limit of these expressions.
Deriving the Limit
The central limit is that increasingly frequent compounding at the fixed annual rate \(r\) approaches the continuous-growth formula.
Here \(e^x\) denotes the natural exponential function. The proof uses the standard limit \(\ln(1+u)/u\to 1\) as \(u\to 0\), which follows from the derivative of \(\ln x\) at \(x=1\).
For \(r\neq 0\) and sufficiently large \(n\), the base is positive, so $$ \ln\left(\left(1+\frac{r}{n}\right)^{nt}\right) =nt\ln\left(1+\frac{r}{n}\right) =rt\,\frac{\ln(1+r/n)}{r/n}. $$
As \(n\to\infty\), \(r/n\to 0\), and therefore $$ rt\,\frac{\ln(1+r/n)}{r/n}\longrightarrow rt. $$
The exponential function is continuous, so the growth factor tends to \(e^{rt}\). Multiplying by \(P_0\) gives the claimed limit. If \(r=0\), every growth factor is \(1\), and the same conclusion holds because \(e^0=1\).
This derivation explains why the exponent contains the product \(rt\). The periodic rate \(r/n\) becomes small as compounding becomes more frequent, but the number of periods \(nt\) becomes large. Their combined effect is captured by the logarithmic limit.
Worked Examples
Worked Example: A Balance Compounded Twice per Year
An initial balance of \(\$1000\) earns a fixed annual rate of \(6\%\). Find the balance after two years if interest is compounded twice per year, then give the continuous-growth limit.
Here \(P_0=1000\), \(r=0.06\), \(t=2\), and \(n=2\). The periodic rate is $$ \frac{r}{n}=\frac{0.06}{2}=0.03, $$ and the number of periods is $$ nt=2\cdot 2=4. $$
The discrete balance is therefore $$ P_2(2)=1000(1.03)^4. $$ Since \(1.03^2=1.0609\), squaring gives \(1.03^4=1.12550881\). Thus $$ P_2(2)=1000(1.12550881)=\$1125.50881, $$ or \(\$1125.51\) to the nearest cent.
For the continuous-growth limit, replace the finite-\(n\) expression by its limit: $$ \lim_{n\to\infty}1000\left(1+\frac{0.06}{n}\right)^{2n} =1000e^{0.06\cdot 2} =1000e^{0.12}. $$ Using \(e^{0.12}\approx 1.127496852\), the limiting balance is approximately \(\$1127.50\) to the nearest cent. The discrete calculation is less than the limit, as expected for this particular finite compounding frequency; it is not itself the limiting value.
As a check on the continuous formula, \(P(t)=1000e^{0.06t}\) gives \(P(0)=1000e^0=1000\), and differentiation gives \(P'(t)=0.06(1000e^{0.06t})=0.06P(t)\).
Worked Example: A Rational Time with Exact Period Counts
Suppose \(P_0=\$400\), the fixed annual rate is \(10\%\), and \(t=3/2\) years. With \(n=2\) compounding periods per year, calculate the discrete balance and the continuous limit.
For \(n=2\), the periodic rate is \(0.10/2=0.05\), and the period count is $$ nt=2\cdot\frac{3}{2}=3. $$ This is an integer, so the time is an exact point on this discrete schedule. The balance is $$ 400(1.05)^3=400(1.05\cdot1.05\cdot1.05) =400(1.157625)=\$463.05. $$
In the limit, the formula gives $$ \lim_{n\to\infty}400\left(1+\frac{0.10}{n}\right)^{3n/2} =400e^{0.10(3/2)} =400e^{0.15}. $$ Since \(e^{0.15}\approx1.161834243\), this is approximately \(\$464.73\) to the nearest cent.
The time \(3/2\) is rational, so there is a sequence of actual schedules whose period counts at this time are integers: take \(n=2,4,6,\ldots\). In each case, \(n(3/2)\) is an integer. The limit along those values of \(n\) agrees with the algebraic limit proved above.
As a check, the limiting function is \(P(t)=400e^{0.10t}\). It has \(P(0)=400\), and \(P'(t)=0.10P(t)\), as required for a continuous relative growth rate of \(0.10\) per year.
Worked Example: An Irrational Time in the Algebraic Limit
Let \(P_0=1\), \(r=0.10\), and \(t=\sqrt{2}\) years. For sufficiently large \(n\), the base \(1+0.10/n\) is positive. The algebraic expression is therefore defined for this real time, and the theorem gives $$ \lim_{n\to\infty}\left(1+\frac{0.10}{n}\right)^{n\sqrt{2}} =e^{0.10\sqrt{2}}. $$
The exponent is \(0.10\sqrt{2}\), not \(0.10n\sqrt{2}\) in the limiting answer: the \(n\) inside the exponent has already been accounted for in the limit. Numerically, \(0.10\sqrt{2}\approx0.141421356\), so the limiting amount is approximately \(1.1519\) to four decimal places.
There is no positive integer \(n\) for which \(n\sqrt{2}\) is an integer: if it were an integer \(m\), then \(\sqrt{2}=m/n\) would be rational. Thus this fixed irrational time is not an exact period-count time for any schedule with an integer number of periods per year. The result here is the algebraic limit at real \(t=\sqrt{2}\), not a claim that one of those schedules has a compounding event exactly at that time.
For a check, taking logarithms of the growth factor gives $$ n\sqrt{2}\ln\left(1+\frac{0.10}{n}\right) =0.10\sqrt{2}\, \frac{\ln(1+0.10/n)}{0.10/n} \longrightarrow 0.10\sqrt{2}. $$ Exponentiating yields exactly the stated limit.
What the Continuous Model Means
The limit \(P(t)=P_0e^{rt}\) is a continuous model derived from a sequence of periodic models. Its rate parameter \(r\) is the fixed annual rate used in the increasingly frequent compounding process. In the continuous model, the balance satisfies $$ P'(t)=rP(t),\qquad P(0)=P_0. $$ Indeed, differentiating \(P_0e^{rt}\) gives \(P'(t)=rP_0e^{rt}=rP(t)\), and substituting \(t=0\) gives \(P(0)=P_0\). This connects the compound-interest limit with the constant-relative-rate differential equation.
The units also match. If \(t\) is measured in years and \(r\) is a rate per year, then \(rt\) is dimensionless, as an exponent must be. The amount \(P_0e^{rt}\) has the same units as \(P_0\).
A Common Mistake
A tempting but incorrect step is to keep only the per-period factor and forget that the number of periods also increases with \(n\). For example, at \(r=0.08\) and \(t=1\), one might incorrectly write $$ \lim_{n\to\infty}\left(1+\frac{0.08}{n}\right)=1 $$ and conclude that the balance does not grow. This expression has only one period’s factor; it omits the \(n\) periods in a year.
The correct expression is $$ \lim_{n\to\infty}\left(1+\frac{0.08}{n}\right)^n=e^{0.08}\approx1.083287, $$ so one unit of initial balance approaches about \(1.083287\) units after one year. The exponent \(n\) is essential: the factor for each period approaches \(1\), while the accumulated effect of all periods approaches \(e^{0.08}\).
When to Interpret the Limit as a Schedule
For a particular \(t\), an actual schedule with \(n\) periods per year lands exactly at time \(t\) only when \(nt\) is an integer. If \(t=a/b\) is rational in lowest terms, choosing \(n\) through positive multiples of \(b\) makes \(nt\) an integer, so the algebraic limit can also be read as a limit along such actual schedules.
For irrational \(t\), no positive integer \(n\) makes \(nt\) an integer. The expression \(P_0(1+r/n)^{nt}\) still has a well-defined algebraic limit for sufficiently large \(n\), because its base is positive and real exponents are defined for positive bases. Its interpretation is then a continuous extension of the discrete formula, rather than a balance recorded exactly at \(t\) by each schedule.
This distinction does not change the continuous-growth formula. It specifies precisely what the limit asserts and prevents the discrete schedule from being confused with its continuous idealization.
Check Your Understanding
Answer these questions using the definitions and limit argument above.
- For \(n\) compounding periods per year, what is the periodic rate and how many periods occur in \(t\) years?
- Use the logarithmic limit to explain why \((1+r/n)^{nt}\) approaches \(e^{rt}\).
- Why is the algebraic expression defined for an irrational time \(t\), even though no integer \(n\) gives an exact period count \(nt\)?
- For \(r=0.04\) and \(t=3\), write the continuous-growth limit factor. Do not evaluate it numerically.
- What is omitted in the incorrect calculation that takes the limit of \(1+r/n\) alone?
