The Exponential Limit (1+x/n)^n

Why This Limit Matters

A sequence can have a limit that is not apparent from any one of its terms. The sequence \((1+1/n)^n\), for example, is built from a base that approaches \(1\), while the exponent grows without bound. Its terms approach neither the base's limit nor a value obtained by simply multiplying the small increment by the exponent.

The limit connects this basic algebraic form to the exponential function. Once the form is recognized, it also provides a reliable way to evaluate sequences whose bases or indices have been rewritten.

Theorem: For every fixed real number \(x\), the sequence \(\left(1+\frac{x}{n}\right)^n\) converges to \(e^x\) as \(n\) tends to infinity through the positive integers.

For any fixed \(x\), the base is positive once \(n>|x|\). Thus, even if the base is negative for some early indices, the sequence is positive from some point onward, which is all that matters for its limit.

The Exponential Function and Its Series

We use the standard power-series representation of the exponential function. In particular, \(e\) is the value of this function at \(1\). For each real \(x\), the series below converges absolutely and represents \(e^x\):

e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!} =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.

Here \(0!=1\), and for a positive integer \(k\), \(k!=1\cdot2\cdots k\). To see why the series converges absolutely, consider the absolute values of its terms. For \(x\ne 0\), the ratio of consecutive terms is \(|x|/(k+1)\), which is at most \(1/2\) for all sufficiently large \(k\). The remaining tail is therefore bounded by a convergent geometric series. For \(x=0\), the series has just one nonzero term.

This representation gives us a way to compare \((1+x/n)^n\) with \(e^x\): expand the power using the binomial theorem and compare the resulting coefficients with those in the exponential series.

Proof of the Exponential Limit

For each positive integer \(n\), the binomial theorem gives a finite sum. Rewriting its coefficients reveals the terms that will converge to those of the exponential series:

\begin{aligned} \left(1+\frac{x}{n}\right)^n &=\sum_{k=0}^{n}\binom{n}{k}\frac{x^k}{n^k}\\ &=\sum_{k=0}^{n}\frac{x^k}{k!} \prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right). \end{aligned}

For \(k=0\), the product is an empty product and has value \(1\). For \(1\le k\le n\), the coefficient identity follows by writing out the binomial coefficient:

\binom{n}{k}\frac{x^k}{n^k} = \frac{x^k}{k!} \frac{n(n-1)\cdots(n-k+1)}{n^k} = \frac{x^k}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right).

Fix \(k\). As \(n\) grows, each of the finitely many factors in the product approaches \(1\). Therefore, for fixed \(k\), the coefficient multiplying \(x^k/k!\) approaches \(1\). To conclude that the whole finite sum approaches the infinite series, we must also control all the terms beyond any fixed \(k\). The absolute convergence established above gives exactly that control.

Choose a finite initial segment.
Given \(\varepsilon>0\), choose an integer \(K\) large enough that \(\sum_{k=K+1}^{\infty}|x|^k/k!<\varepsilon/4\).

Control the first \(K+1\) terms.
For \(n\ge K\), each of the finitely many products \(\prod_{j=0}^{k-1}(1-j/n)\), for \(0\le k\le K\), approaches \(1\). Hence, for all sufficiently large \(n\), the absolute difference between the first \(K+1\) terms of the binomial expansion and the first \(K+1\) terms of the exponential series is less than \(\varepsilon/2\).

Bound both tails.
For \(0\le j\le k-1<n\), each factor \(1-j/n\) lies between \(0\) and \(1\). Thus the absolute value of the binomial term for index \(k\) is at most \(|x|^k/k!\). The binomial tail and the exponential-series tail are each less than \(\varepsilon/4\).

Combine the estimates.
The total absolute difference is less than \(\varepsilon/2+\varepsilon/4+\varepsilon/4=\varepsilon\), for all sufficiently large \(n\). This proves that \((1+x/n)^n\) converges to \(\sum_{k=0}^{\infty}x^k/k!=e^x\).

The argument explains why the limit holds: each fixed binomial coefficient approaches the corresponding exponential-series coefficient, and the remaining terms are controlled by the convergent series of absolute values. It applies to negative \(x\) as well as positive \(x\), since the tail estimates use absolute values.

Worked Examples

Worked Example: A Negative Fixed Value

Find the limit of \(\left(1-\frac{2}{n}\right)^n\). This is the theorem's form with \(x=-2\), so the predicted limit is \(e^{-2}\). The base is positive for every \(n>2\), and therefore is positive for all sufficiently large \(n\), as required for the limiting behavior.

\begin{aligned} \lim_{n\to\infty}\left(1-\frac{2}{n}\right)^n &=\lim_{n\to\infty}\left(1+\frac{-2}{n}\right)^n\\ &=e^{-2}. \end{aligned}

Check: the expression has exactly the form \((1+x/n)^n\), with the fixed value \(x=-2\). Substituting that value into the theorem gives \(e^x=e^{-2}\), confirming the result.

Worked Example: A Multiple of the Index

Find the limit of \(\left(1+\frac{6}{3n}\right)^{3n}\). The denominator and the exponent both involve \(3n\), so set \(m=3n\). As \(n\) tends to infinity through the positive integers, \(m\) also tends to infinity through the positive integers. The expression becomes the theorem's form with index \(m\) and fixed value \(6\):

\begin{aligned} \left(1+\frac{6}{3n}\right)^{3n} &=\left(1+\frac{6}{m}\right)^m \qquad (m=3n),\\ \lim_{n\to\infty}\left(1+\frac{6}{3n}\right)^{3n} &=e^6. \end{aligned}

Check: substituting \(m=3n\) back into \((1+6/m)^m\) gives \((1+6/(3n))^{3n}\), the original sequence. The fixed value in the numerator is \(6\), not \(6/3\), because the full denominator is \(3n\).

Worked Example: A Shifted Denominator and Exponent

Find the limit of \(\left(1+\frac{2}{n+1}\right)^n\). The exponent is one less than the denominator's index, so first use the exponential limit on the closely related power with exponent \(n+1\). Then express the original sequence as a quotient.

\begin{aligned} A_n &=\left(1+\frac{2}{n+1}\right)^{n+1} \longrightarrow e^2,\\ \left(1+\frac{2}{n+1}\right)^n &=\frac{\left(1+\frac{2}{n+1}\right)^{n+1}} {1+\frac{2}{n+1}} =\frac{A_n}{1+\frac{2}{n+1}}. \end{aligned}

As \(n\) grows, \(2/(n+1)\) tends to \(0\), so the denominator \(1+2/(n+1)\) tends to \(1\). By the quotient limit law, whose denominator limit is nonzero, the original sequence has limit \(e^2/1=e^2\).

\lim_{n\to\infty}\left(1+\frac{2}{n+1}\right)^n = \frac{e^2}{1} =e^2.

Check: multiplying the quotient by the removed factor \(1+2/(n+1)\) restores the power \(n+1\), so the algebraic identity used in the calculation is exact.

Recognizing Equivalent Forms

The key is to identify the index that appears in both the small increment and the exponent. A constant multiple of \(n\) can serve as that index. A shifted index may also be useful, as in the third example. Once the expression is rewritten, identify the fixed number playing the role of \(x\).

Expression	Useful index	Limit
\(\left(1+\frac{4}{n}\right)^n\)	\(n\)	\(e^4\)
\(\left(1+\frac{5}{2n}\right)^{2n}\)	\(m=2n\)	\(e^5\)
\(\left(1+\frac{3}{n+1}\right)^n\)	Rewrite using exponent \(n+1\), then divide by the base	\(e^3\)

In the second row, setting \(m=2n\) gives \((1+5/m)^m\). In the third row, the exact identity is \((1+3/(n+1))^n=(1+3/(n+1))^{n+1}/(1+3/(n+1))\). The limit in each case follows from the theorem and, in the shifted case, the quotient limit law.

A Common Mistake

A tempting but incorrect calculation is to expand the power and keep only the constant and linear terms, as though all higher terms vanish just because \(x/n\) tends to zero. For \(x=1\), this mistake gives:

\left(1+\frac{1}{n}\right)^n \mathrel{\not\approx} 1+n\left(\frac{1}{n}\right)=2 \quad\text{as a limit calculation.}

The binomial expansion actually contains terms of every degree up to \(n\). For each fixed \(k\), the term of degree \(k\) approaches \(1/k!\), not \(0\). The higher-degree terms collectively contribute to the limit:

\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \sum_{k=0}^{\infty}\frac{1}{k!} =e, \qquad\text{not }2.

The correction is not to retain just one or two terms, but to account for the full series and control its tail, as in the proof.

Why the Result Is Useful

This limit is a standard pattern for recognizing the exponential function inside a sequence. It is particularly useful when algebra disguises the index, or when a simple factorization reveals a power with a shifted exponent. The proof also illustrates an important principle in sequence limits: convergence of each fixed term in an expansion is not enough by itself; the remaining terms must be controlled as well.

Takeaway: If an expression can be rewritten exactly as \(\left(1+\frac{x}{m}\right)^m\), where \(x\) is fixed and the integer index \(m\) tends to infinity, its limit is \(e^x\). For shifted exponents, first rewrite the expression exactly and then use the limit laws that apply.

Check Your Understanding

Use the theorem and exact algebraic rewrites where needed. State the fixed value and index used in each application.

What is the limit of \(\left(1+\frac{3}{n}\right)^n\)?
Rewrite \(\left(1+\frac{8}{4n}\right)^{4n}\) using an index \(m\), then find its limit.
Find the limit of \(\left(1-\frac{5}{n}\right)^n\). For which positive integers \(n\) is its base positive?
Express \(\left(1+\frac{4}{n+1}\right)^n\) as a quotient involving a power with exponent \(n+1\), and find its limit.
Why does retaining only the constant and linear terms in the binomial expansion fail to establish the limit?