The Compound-Growth Limit Defining e

Compound Growth and the Constant e

Suppose an amount grows by a small percentage repeatedly. The number of growth periods matters: dividing a fixed total growth into more frequent steps changes the resulting amount. A standard model for one unit of total growth, divided into \(n\) equal periods, is \(\left(1+\frac{1}{n}\right)^n\).

As the number of periods increases, these values approach a fixed constant. That constant is \(e\), a fundamental number in calculus. To use it carefully, we first establish that the limiting value exists and then show that the same value describes a limit with a real variable.

Establishing the Compound-Growth Constant

For each positive integer \(n\), define \(a_n=\left(1+\frac{1}{n}\right)^n\). The binomial theorem gives a useful expression for this sequence:

For \(k=0\), the product is understood to be \(1\). For a fixed \(k\), every factor in the product is positive when \(n\ge k\), and the factor \(1-\frac{j}{n}\) does not decrease as \(n\) increases. Thus, when comparing \(a_n\) and \(a_{n+1}\), the terms with \(0\le k\le n\) do not decrease, and \(a_{n+1}\) also has one additional positive term. Therefore \(a_{n+1}>a_n\).

The same expansion gives an upper bound. For \(k\ge 1\), each product factor is at most \(1\), so the \(k\)-th term is at most \(1/k!\). Also, \(k!\ge 2^{k-1}\) for \(k\ge1\): the factors \(2,3,\ldots,k\) are each at least \(2\). Consequently, the terms from \(k=2\) onward are bounded by a geometric sum:

The sequence is increasing and bounded above, so it converges by the monotone convergence property of the real numbers. This establishes a finite limiting value; we will show that it is also the limit of the corresponding expression for real exponents tending to infinity.

From Integer Periods to Real Exponents

For real \(t>0\), let \(F(t)=\left(1+\frac{1}{t}\right)^t\). We prove that \(F(t)\) approaches the same limit as \(a_n\). The bounding argument below is used only for \(t\ge1\), which is enough to determine the limit as \(t\to\infty\).

Choose the integer \(m\) such that \(m\le t<m+1\). Since \(t\ge1\), we have \(m\ge1\), so all the expressions below are defined. The base decreases as \(t\) increases, whereas the exponent increases. We therefore bound the base and exponent in the directions that give valid bounds for the whole power. Since all bases are greater than \(1\),

The lower base raised to \(m\) is no larger than the actual base raised to \(t\): the base is no larger, and \(t\ge m\). The actual base raised to \(t\) is no larger than the upper base raised to \(m+1\): the base is no larger, and \(t<m+1\). Hence

Both bounds have limits given by the sequence \(a_n\), with factors tending to \(1\):

As \(t\to\infty\), the integer \(m\) also tends to infinity. The squeeze theorem now shows that \(F(t)\) has the same limit as \(a_n\).

The Two-Sided Increment Limit

To study increments close to zero, set \(t=1/h\) when \(h>0\). Then \(t\to\infty\) as \(h\to0^+\), and

For the negative side, write \(h=-u\), where \(u>0\) and \(u\to0^+\). Set \(t=\frac{1-u}{u}\), which tends to infinity. Since \(1+\frac{1}{t}=\frac{1}{1-u}\) and \(t+1=\frac{1}{u}\), we obtain

The original expression requires \(1+h>0\), so its real-valued domain includes all \(h\) sufficiently close to zero. The positive- and negative-side limits agree, giving the two-sided result.

Worked Examples with Scaled Increments

From the Compound-Growth Limit to the Exponential Increment

The limit just proved also yields the increment limit that appears in the difference quotient for the natural exponential. We use the standard inverse relationship between the natural logarithm and exponential: for \(y>0\), \(\ln y\) is the number \(v\) for which \(e^v=y\). The logarithm is continuous on positive inputs, and the exponential is continuous, strictly increasing, and satisfies \(e^0=1\).

Apply the natural logarithm to \((1+h)^{1/h}\). For \(h\ne0\) close to zero, \(1+h>0\), and the logarithm power rule gives

The left side tends to \(\ln e=1\), by the compound-growth limit and continuity of the logarithm. Thus \(\frac{\ln(1+h)}{h}\to1\) as \(h\to0\). Now set \(h=e^u-1\). As \(u\to0\), continuity of the exponential gives \(h\to0\). Also, \(\ln(1+h)=\ln(e^u)=u\). For \(u\ne0\), strict increase of the exponential ensures \(e^u-1\ne0\), and so

Taking reciprocals is valid because the displayed quotient tends to the nonzero value \(1\). We conclude that

A Common Scaling Pitfall

A frequent error is to change the increment in the base without changing the exponent accordingly. For example, for \(\lim_{x\to0}(1+2x)^{1/x}\), setting \(h=2x\) and then writing the expression as \((1+h)^{1/h}\) is incorrect: \(1/h=1/(2x)\), not \(1/x\). That incorrect substitution would give \(e\).

The correct exponent conversion is \(1/x=2/h\), so the expression is \((1+h)^{2/h}\). Therefore

Always convert both the base and exponent when introducing a new increment variable. The limit \((1+h)^{1/h}\to e\) applies directly only when the exponent is exactly \(1/h\).

Why These Limits Matter

Compound growth gives a concrete way to define \(e\), while the increment limit translates that definition into information about how \(e^u\) changes near \(u=0\). In differentiation, difference quotients often contain increments of a function. The limit \(\frac{e^u-1}{u}\to1\) supplies the precise local behavior needed for the natural exponential.

The essential checks are the domain of the base, the direction from which the increment approaches zero, and the exact exponent after substitution. Keeping those conditions visible prevents both invalid limit arguments and missing scaling factors.