Show, using a table of values, that $\lim_{n o \infty} \left(1+\dfrac1n ight)^n = e \approx 2.71828$.

Question

Show, using a table of values, that $\lim_{n 	o \infty} \left(1+\dfrac1night)^n = e \approx 2.71828$.

Accepted Answer

Unlike the other standard limits, $\lim_{n 	o \infty}\left(1+\frac1night)^n$ is not found by substitution or factorisation — the value $e$ is, by definition, whatever this expression settles down to as $n$ grows without bound. This is best seen by computing the expression for successively larger values of $n$:

| $n$ | $\left(1+\frac1night)^n$ |
|---|---|
| $10$ | $2.5937$ |
| $100$ | $2.7048$ |
| $1{,}000$ | $2.7169$ |
| $10{,}000$ | $2.7181$ |
| $100{,}000$ | $2.71827$ |

As $n$ increases, the value keeps rising but by smaller and smaller amounts, closing in on the fixed number $$e \approx 2.71828\ldots$$ (an irrational number, meaning its decimal expansion never terminates or repeats). This is exactly how $e$ is defined in the first place: $$e = \lim_{n 	o \infty}\left(1+\dfrac1night)^n$$ …

Q.Show, using a table of values, that $\lim_{n \to \infty} \left(1+\dfrac1n\right)^n = e \approx 2.71828$ .

Concept understanding — Standard Limits

A Toolkit of Named Limits

The Sandwich Theorem

The two flagship trigonometric limits

The full standard-limit toolkit (§9.2.10)

The recognise-and-substitute pattern

$n$	$\left(1+\frac1n\right)^n$
$10$	$2.5937$
$100$	$2.7048$
$1{,}000$	$2.7169$
$10{,}000$	$2.7181$
$100{,}000$	$2.71827$

Q.Show, using a table of values, that lim⁡n→∞(1+1n)n=e≈2.71828\lim_{n \to \infty} \left(1+\dfrac1n\right)^n = e \approx 2.71828limn→∞​(1+n1​)n=e≈2.71828.

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Q.Show, using a table of values, that $\lim_{n \to \infty} \left(1+\dfrac1n\right)^n = e \approx 2.71828$ .

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