Evaluate lim(n→∞) (1 + 1/n)^n = e from first principles

Question

Show that $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$ . Explain why this limit exists and what it converges to.

(JEE Main 2022, conceptual)

Solution — Step by Step

For positive integer $n$ :

\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k}

= 1 + n \cdot \frac{1}{n} + \frac{n(n-1)}{2!} \cdot \frac{1}{n^2} + \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{n^3} + \cdots

= 1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \cdots

As $n \to \infty$ , each factor $\left(1 - \frac{k}{n}\right) \to 1$ .

\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots = \sum_{k=0}^{\infty} \frac{1}{k!}

This infinite series is precisely the definition of $e$ .

e = 1 + 1 + 0.5 + 0.1667 + 0.0417 + 0.0083 + \cdots \approx 2.71828

The series converges rapidly because $k!$ grows very fast.

Therefore, $\boxed{\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \approx 2.71828}$

Why This Works

The expression $(1 + 1/n)^n$ represents compound interest taken to its logical extreme. If a bank offers 100% annual interest compounded $n$ times per year, after one year you’d have $(1 + 1/n)^n$ times your initial deposit. As $n \to \infty$ (continuous compounding), this approaches $e$ — roughly 2.718 times your deposit.

The sequence is monotonically increasing and bounded above (by the series sum), so it must converge. The limit is called Euler’s number $e$ , one of the most important constants in mathematics — it appears in calculus, probability, and complex analysis.

Alternative Method — Using the logarithm

Let $L = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$ .

Take $\ln$ : $\ln L = \lim_{n \to \infty} n \ln\left(1 + \frac{1}{n}\right)$ .

Let $x = 1/n$ , so as $n \to \infty$ , $x \to 0$ :

\ln L = \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1

(using the standard limit or L’Hopital’s rule)

So $\ln L = 1$ , giving $L = e^1 = e$ .

The generalised form $\lim_{n \to \infty} \left(1 + \frac{a}{n}\right)^n = e^a$ is extremely useful in JEE. Any limit of the form $1^\infty$ can often be rewritten in this form. The shortcut: if $\lim f(x) = 1$ and $\lim g(x) = \infty$ , then $\lim [f(x)]^{g(x)} = e^{\lim g(x)[f(x) - 1]}$ .

Common Mistake

Students sometimes claim $(1 + 1/n)^n \to 1$ because ” $1$ raised to anything is $1$ .” This is the classic $1^\infty$ trap. The base $1 + 1/n$ is NOT exactly $1$ — it’s slightly greater than $1$ , and as $n$ grows, we raise it to a larger and larger power. The “slightly more than 1” and “very large exponent” compete, and the result is a finite number ( $e$ ), not $1$ or $\infty$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the logarithm

Common Mistake

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