The number e — definition as limit of (1+1 - n)ⁿ, natural growth context: Exercises

New here? Read the parent note first — it builds every symbol below from zero.

Definition Two tools this page keeps using — defined once, here

The natural logarithm $ln$ . The exponential $e^{x}$ takes an exponent and returns a growth factor. Its inverse, the natural logarithm $ln$ , runs backwards: $ln y$ answers the question "what exponent do I feed $e$ to get $y$ ?" In symbols, $ln y = c$ means exactly $e^{c} = y$ . Its base is $e$ itself (that's what "natural" means — see Natural logarithm ln x). Two facts we lean on: it undoes $e$ , so $ln (e^{x}) = x$ ; and it turns powers into products, $ln (a^{b}) = b ln a$ . Useful values: $ln 1 = 0$ , $ln 2 = 0.693147$ , $ln 3 = 1.098612$ .

The "size" shorthand $O (\cdot)$ . When we write a quantity is $O (\frac{1}{n ^{2}})$ , we mean plainly: its magnitude is at most some fixed constant times $\frac{1}{n ^{2}}$ once $n$ is large. It is a book-keeping label for "a leftover that shrinks at least this fast." The point is comparison: $O (\frac{1}{n ^{2}})$ vanishes faster than $O (\frac{1}{n})$ , which vanishes faster than a constant. We only ever use it to say "this leftover dies quickly enough to ignore in the limit."

Level 1 — Recognition

Goal: recognise the definition and read it correctly, no computation dexterity required.

Recall Solution 1.1

(a). By definition $e = lim_{n \to \infty} (1 + \frac{1}{n})^{n}$ — the whole limit statement, in which the base $(1 + \frac{1}{n})$ tends to $1$ and the power $n$ tends to $\infty$ together. Both moving parts must be present inside the same limit.

(b) keeps the base but drops the power, so it just tends to $1 + 0 = 1$ . That is not $e$ .
(c) has the pieces swapped: base $\to \infty$ , power $\to 0$ . This is the $\infty^{0}$ form. WHY it equals $1$ , not $\infty$ : take logs — the log of $(1 + n)^{1/ n}$ is $ln [(1 + n)^{1/ n}] = \frac{l n ( 1 + n )}{n} .$ The top grows like $ln n$ (very slow), the bottom grows like $n$ (fast), so the ratio $\to 0$ . Since the log tends to $0$ , the quantity itself tends to $e^{0} = 1$ . So (c) is $1$ , not $e$ .

Answer: (a).

Recall Solution 1.2

False. For finite $n$ you get a number below $e$ ; only the limit $n \to \infty$ hits $e$ . Check $n = 5$ : $(1 + \frac{1}{5})^{5} = 1. 2^{5} = 2.48832,$ which is clearly less than $2.71828$ . The value only approaches $e$ as $n$ grows.

Recall Solution 1.3

Set $x = 3$ : $e^{3} = lim_{n \to \infty} (1 + \frac{3}{n})^{n} .$ The $x$ sits in the numerator of the chunk $\frac{x}{n}$ — the "total growth" being spread over $n$ steps. WHY this really is $e^{3}$ and not $1$ : by the log trick above, $ln [(1 + \frac{3}{n})^{n}] = n ln (1 + \frac{3}{n}) \approx n \cdot \frac{3}{n} = 3$ , so the quantity tends to $e^{3}$ . The base tending to $1$ does not win — the exponent compounds the surplus $\frac{3}{n}$ exactly enough to reach exponent $3$ .

Level 2 — Application

Goal: plug numbers into $e^{x} = lim (1 + x / n)^{n}$ , $A = P e^{r t}$ , $N = N_{0} e^{k t}$ .

Recall Solution 2.1

$n = 4 : (1.25)^{4} = 2.44141, n = 8 : (1.125)^{8} = 2.56578.$ The sequence increases toward $e$ , so $n = 8$ (value $2.56578$ ) is closer to $2.71828$ .

Figure — The number e — definition as limit of (1+1 - n)ⁿ, natural growth context

What to observe in the figure: the horizontal axis is $n$ , the vertical axis is $(1 + 1/ n)^{n}$ . The blue dots (the sequence) climb steadily; the dashed yellow line is the ceiling $e \approx 2.71828$ . Notice the pink squares at $n = 4$ and $n = 8$ : the second is visibly higher and nearer the yellow line, confirming $n = 8$ is closer. Crucially, the dots never cross the yellow line — the sequence rises but is trapped below $e$ (the monotone-bounded picture proved in Ex 3.2).

Recall Solution 2.2

Use $A = P e^{r t}$ with $P = 2000$ , $r = 0.04$ , $t = 5$ . WHAT: the exponent is the total accumulated growth $r t = 0.04 \times 5 = 0.20$ . WHY: $A = P e^{r t}$ is the $n \to \infty$ limit of $P (1 + \frac{r}{n})^{n t}$ — see Compound interest. $A = 2000 e^{0.20} = 2000 \times 1.221403 = £2442.81.$

Recall Solution 2.3

Continuous growth solves $\frac{d N}{d t} = k N$ , giving $N = N_{0} e^{k t}$ (see Differential equations dy/dx = ky). $N = 800 e^{0.3 \times 6} = 800 e^{1.8} = 800 \times 6.049647 = 4839.7 \approx 4840 cells.$ The exponent $k t = 1.8$ is the total growth accumulated over the $6$ hours.

Recall Solution 2.4

First five terms ( $k = 0, 1, 2, 3, 4$ ): $1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} = 1 + 1 + 0.5 + 0.166667 + 0.041667 = 2.708333.$ The true value is $2.718282$ , so the error is $2.718282 - 2.708333 = 0.009949 \approx 0.00995$ . The remaining terms ( $\frac{1}{5 !} + \dots$ ) fill this small gap — factorials shrink them fast.

Level 3 — Analysis

Goal: explain WHY the machinery behaves as it does — convergence, boundedness, the indeterminate tug-of-war.

Recall Solution 3.1

Write $e = 1 + 1 + \frac{1}{2 !} + \frac{1}{3 !} + \frac{1}{4 !} + \dots$ . Key comparison: for $k \geq 2$ , $k! = 2 \cdot 3 \cdot 4 \dots k \geq 2 \cdot 2 \cdot 2 \dots 2 = 2^{k - 1}$ , so $\frac{1}{k !} \leq \frac{1}{2 ^{k - 1}}$ . Therefore $e = 1 + 1 + \sum_{k \geq 2} \frac{1}{k !} < 1 + geometric, sums to 2 (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots) = 1 + 2 = 3.$ The inequality is strict because from $k = 3$ onward $k! > 2^{k - 1}$ (e.g. $3! = 6 > 4$ ), so at least one term is strictly smaller than its geometric bound. Hence $e < 3$ . This is why the greedy "infinite money" intuition fails — the corrections are crushed by factorials.

What to observe in the figure: the pink bars are the individual series terms $\frac{1}{k !}$ against $k$ — see how fast they collapse toward zero after $k = 2$ . The blue running-total curve climbs and then flattens, pinned under the yellow line $e$ . The dotted white line marks the bound $3$ : the total never reaches it. The picture is the proof — shrinking bars ⇒ bounded sum.

Recall Solution 3.2

Let $a_{n} = (1 + \frac{1}{n})^{n}$ and $b_{n} = (1 + \frac{1}{n})^{n + 1}$ .

WHY $a_{n}$ increases — check the ratio $a_{n + 1} / a_{n} > 1$ . The cleanest tool is the AM–GM inequality (the average of numbers is at least their geometric mean). Take $n$ copies of $(1 + \frac{1}{n})$ and one copy of $1$ ; their arithmetic mean is $\frac{n ( 1 + \frac{1}{n} ) + 1}{n + 1} = \frac{n + 1 + 1}{n + 1} = 1 + \frac{1}{n + 1} .$ AM–GM says this mean is $\geq$ the geometric mean $[(1 + \frac{1}{n})^{n} \cdot 1]^{1/ (n + 1)}$ . Raising both sides to the power $n + 1$ : $(1 + \frac{1}{n + 1})^{n + 1} \geq (1 + \frac{1}{n})^{n}, i.e. a_{n + 1} \geq a_{n} .$ So $a_{n}$ increases — each extra compounding chunk genuinely adds value.

WHY $b_{n}$ decreases — the explicit AM–GM on reciprocals. Look at $\frac{1}{b _{n}} = (\frac{n}{n + 1})^{n + 1} = (1 - \frac{1}{n + 1})^{n + 1}$ ; if this increases then $b_{n}$ decreases. Take $n + 1$ copies of $(1 - \frac{1}{n + 1}) = \frac{n}{n + 1}$ and one copy of $1$ . Their arithmetic mean is $\frac{( n + 1 ) \cdot \frac{n}{n + 1} + 1}{n + 2} = \frac{n + 1}{n + 2} = 1 - \frac{1}{n + 2} .$ AM–GM gives $1 - \frac{1}{n + 2} \geq [(1 - \frac{1}{n + 1})^{n + 1} \cdot 1]^{1/ (n + 2)}$ . Raising to the power $n + 2$ : $(1 - \frac{1}{n + 2})^{n + 2} \geq (1 - \frac{1}{n + 1})^{n + 1}, i.e. \frac{1}{b _{n + 1}} \geq \frac{1}{b _{n}} .$ Taking reciprocals flips the inequality: $b_{n + 1} \leq b_{n}$ . So $b_{n}$ decreases toward $e$ from above.

Why $e$ is trapped. Since $b_{n} = a_{n} \cdot (1 + \frac{1}{n}) > a_{n}$ , and $a_{n}$ climbs while $b_{n}$ falls, we get $a_{n} < e < b_{n}$ for every $n$ . The gap $b_{n} - a_{n} = a_{n} \cdot \frac{1}{n} \to 0$ , so both are squeezed onto the same limit $e$ .

Numeric check at $n = 10$ : $a_{10} = 2.593742$ , $b_{10} = 2.853116$ , and indeed $a_{10} < e < b_{10}$ .

Recall Solution 3.3

Step 1 — each term's limit. Each bracketed factor $(1 - \frac{j}{n})$ tends to $1 - 0 = 1$ as $n \to \infty$ , because $\frac{j}{n} \to 0$ for fixed $j$ . There are finitely many such factors ( $k - 1$ of them), so their product tends to $1 \cdot 1 \dots 1 = 1$ . Hence $T_{k} \to \frac{1}{k !}$ .

Step 2 — why the whole sum follows the terms (a plain squeeze). We want $(1 + \frac{1}{n})^{n} = \sum_{k = 0}^{n} T_{k}$ to tend to $\sum_{k = 0}^{\infty} \frac{1}{k !}$ . Here is an elementary argument, no advanced machinery.

Upper bound: since every bracket $(1 - \frac{j}{n}) \leq 1$ , we have $T_{k} \leq \frac{1}{k !}$ , so $(1 + \frac{1}{n})^{n} = \sum_{k = 0}^{n} T_{k} \leq \sum_{k = 0}^{n} \frac{1}{k !} \leq e$ . So the sequence never exceeds $e$ .
Lower bound: fix any whole number $M$ . For $n > M$ , keep only the first $M + 1$ terms (dropping the rest, which are positive): $(1 + \frac{1}{n})^{n} \geq \sum_{k = 0}^{M} T_{k}$ . Now let $n \to \infty$ with $M$ held fixed — by Step 1 each of these finitely many $T_{k} \to \frac{1}{k !}$ , so the right side $\to \sum_{k = 0}^{M} \frac{1}{k !}$ . Hence $lim inf_{n} (1 + \frac{1}{n})^{n} \geq \sum_{k = 0}^{M} \frac{1}{k !}$ .
Squeeze: the lower bound holds for every $M$ , so letting $M \to \infty$ gives $lim inf \geq \sum_{k = 0}^{\infty} \frac{1}{k !} = e$ . Combined with the upper bound $\leq e$ , the limit is exactly $e$ .

That is how the limit becomes the series — bounded above by the tail-sum and pushed up from below by longer and longer initial chunks.

Level 4 — Synthesis

Goal: combine the limit, the series, exponential growth, and its inverse across several steps.

Recall Solution 4.1

We need the growth factor $e^{0.06 t} = 2$ . Why $ln$ ? $ln$ is the inverse of $e^{x}$ — it answers "what exponent produces this factor?" Apply it to undo the exponential: $0.06 t = ln 2 ⟹ t = \frac{l n 2}{0.06} = \frac{0.693147}{0.06} = 11.5525 years .$ So about $11.55$ years to double. (This is the continuous-growth "rule of 69.3".)

Recall Solution 4.2

$N_{A} = N_{0} e^{0.10 t}$ , $N_{B} = N_{0} e^{0.25 t}$ . Require $N_{B} = 3 N_{A}$ : $N_{0} e^{0.25 t} = 3 N_{0} e^{0.10 t} ⟹ e^{0.25 t - 0.10 t} = 3 ⟹ e^{0.15 t} = 3.$ The $N_{0}$ cancels (both started equal). Take $ln$ : $0.15 t = ln 3 ⟹ t = \frac{l n 3}{0.15} = \frac{1.098612}{0.15} = 7.3241 hours .$ About $7.32$ hours.

Recall Solution 4.3

Step 1 — the honest product, with its cross-term. Multiply the two brackets: $(1 + \frac{x}{n}) (1 + \frac{y}{n}) = 1 + \frac{x + y}{n} + cross-term \frac{x y}{n ^{2}} .$ The last piece $\frac{x y}{n ^{2}}$ is $O (\frac{1}{n ^{2}})$ (recall the top-of-page shorthand: a leftover no bigger than a constant times $\frac{1}{n ^{2}}$ ) — it shrinks faster than the $\frac{x + y}{n}$ term that drives the growth. So for large $n$ , $(1 + \frac{x}{n}) (1 + \frac{y}{n}) = 1 + \frac{x + y}{n} + O (\frac{1}{n ^{2}}) .$

Step 2 — raise to the power $n$ and take logs. $ln [(1 + \frac{x}{n}) (1 + \frac{y}{n})]^{n} = n ln (1 + \frac{x + y}{n} + O (\frac{1}{n ^{2}})) .$ Using $ln (1 + u) \approx u$ for small $u = \frac{x + y}{n} + O (\frac{1}{n ^{2}})$ : $= n (\frac{x + y}{n} + O (\frac{1}{n ^{2}})) = x + y + = O (1/ n) \to 0 n \cdot O (\frac{1}{n ^{2}}) \to x + y .$ Why the cross-term dies: multiplying an $O (\frac{1}{n ^{2}})$ leftover by the outer $n$ leaves only $O (\frac{1}{n})$ , which tends to $0$ . So the log of the product-raised-to- $n$ tends to $x + y$ , meaning the product itself tends to $e^{x + y}$ .

Step 3 — split it as $e^{x} \cdot e^{y}$ (product of limits). Separately $(1 + \frac{x}{n})^{n} \to e^{x}$ and $(1 + \frac{y}{n})^{n} \to e^{y}$ , and both limits exist and are finite. The limit of a product equals the product of the limits precisely when each limit exists finitely — which it does here. Therefore $e^{x} \cdot e^{y} = lim_{n} (1 + \frac{x}{n})^{n} \cdot lim_{n} (1 + \frac{y}{n})^{n} = lim_{n} [(1 + \frac{x}{n}) (1 + \frac{y}{n})]^{n} = e^{x + y} .$ The middle equality is the product-of-limits law; the last equality is Step 2. Growth factors multiply because growth compounds.

Numeric confirmation ( $x = 0.5$ , $y = 1.5$ ): $e^{0.5} = 1.648721$ , $e^{1.5} = 4.481689$ ; product $= 7.389056$ . And $e^{0.5 + 1.5} = e^{2} = 7.389056$ . ✓

Level 5 — Mastery

Goal: push to the edge — variable exponents, zero/negative rates, and limiting behaviour.

Recall Solution 5.1

Reshape to the known template $(1 + \frac{x}{m})^{m} \to e^{x}$ . Let $m = n$ ; the exponent is $3 n = 3 \cdot n$ , so $(1 + \frac{2}{n})^{3 n} = [(1 + \frac{2}{n})^{n}]^{3} \to (e^{2})^{3} = e^{6} .$ WHY this works: the inner bracket matches $x = 2$ (limit $e^{2}$ ); the outer power $3$ is a fixed exponent that survives the limit. Numerically $e^{6} = 403.4288$ .

Recall Solution 5.2

(a) Zero rate. With $x = 0$ the base is exactly $1$ for every $n$ , so $1^{n} = 1$ for all $n$ and the limit is $1$ . No indeterminacy here — the base is literally $1$ , not merely tending to it. This matches $e^{0} = 1$ : no growth, no change. (b) Negative rate. This is $(1 + \frac{- 1}{n})^{n}$ with $x = - 1$ , so the limit is $e^{- 1} = \frac{1}{e} = 0.367879$ . A subtlety about negative $x$ : the formula $e^{x} = lim_{n \to \infty} (1 + \frac{x}{n})^{n}$ needs the base $1 + \frac{x}{n}$ to stay positive so the power is well defined — which requires $n > - x$ . For any fixed negative $x$ this holds for all large enough $n$ (here $n > 1$ ), and since limits only care about large $n$ , the formula still gives $e^{x}$ . WHY it shrinks: a negative exponent means continuous decay — each chunk multiplies by something below $1$ . This is the decay branch of $\frac{dN}{dt}=kN$ with $k < 0$ .

What to observe in the figure: the horizontal axis is the exponent $x$ , the vertical axis the value. The yellow curve $e^{x}$ (for $x > 0$ ) shoots upward — growth. The blue curve $e^{- x}$ (for $x < 0$ ) sinks toward zero — decay. Both pass through the pink dot at $x = 0$ , value $1$ : the zero-rate case (a), where nothing changes. The blue square marks part (b): $(1 - 1/ n)^{n} \to e^{- 1} = 0.368$ . One picture holds all three regimes — grow, stay, decay.

Recall Solution 5.3

Fraction remaining: $\frac{N}{N _{0}} = e^{- 0.05 \times 30} = e^{- 1.5} = 0.223130$ , i.e. about $22.3%$ . Half-life $t_{1/2}$ : solve $e^{- λ t} = \frac{1}{2}$ . Apply $ln$ (inverse of $e^{x}$ ): $- λ t = ln \frac{1}{2} = - ln 2 ⟹ t_{1/2} = \frac{l n 2}{λ} = \frac{0.693147}{0.05} = 13.8629 days .$ About $13.86$ days.

Recall Solution 5.4

In (a) with $x = 0$ the base equals $1$ exactly — there is nothing for the exponent to amplify, so the answer is a plain $1$ (not indeterminate). With $x = 1$ the base is $1 + \frac{1}{n}$ , which is strictly greater than $1$ for every finite $n$ . Take the log: $n ln (1 + \frac{1}{n}) \approx n \cdot \frac{1}{n} = 1$ , which tends to $1$ , not $0$ — so the quantity tends to $e^{1} = e$ . The surplus $\frac{1}{n}$ shrinks like $\frac{1}{n}$ while the repetitions grow like $n$ ; their product stays at $1$ . That is the genuine $1^{\infty}$ tug-of-war. The difference is: "tends to $1$ " is not the same as "equals $1$ ." Only when the base is truly $1$ does the power become irrelevant.

Active recall

Recall Quick self-check (open after attempting)

Which limit is $e$ : $(1 + \frac{1}{n})^{n}$ or $(1 + \frac{1}{n})$ ? → the first (base and power both move).
$(1 + \frac{2}{n})^{3 n} \to$ ? → $e^{6}$ .
$(1 - \frac{1}{n})^{n} \to$ ? → $e^{- 1}$ .
Why does $(1 + 0/ n)^{n} = 1$ but $(1 + 1/ n)^{n} = e$ ? → exact $1$ vs. approaching $1$ ; the log of the second is $n ln (1 + \frac{1}{n}) \to 1$ , not $0$ .
Continuous doubling time at rate $r$ ? → $t = \frac{l n 2}{r}$ .

Connections

The number $e$ — definition as limit of $(1+1/n)^n$, natural growth context — the parent this drills.
Natural logarithm ln x — the inverse used to solve for $t$ in L4/L5.
Exponential function e^x and its derivative — the function behind $A = P e^{r t}$ .
Compound interest — origin of Ex 2.2.
Differential equations dy/dx = ky — growth (Ex 2.3) and decay (Ex 5.3).
Binomial theorem — Ex 3.3's expansion.
Limits and indeterminate forms — the $1^{\infty}$ analysis in L1/L5.