Applications — approximation, evaluating limits: Worked examples

The scenario matrix

Every problem in this topic lands in exactly one of nine boxes — Figure s02 shows them as a map, so you can see which cell your problem sits in rather than read a heavy table.

Figure — Applications — approximation, evaluating limits

Figure s02 — the scenario map. Top row (blue): the three approximation cells A/B/C, arranged left-to-right by how far the input sits from the expansion center (small $+$ , small $-$ , edge of convergence). Bottom rows (green): the six limit cells D–I, from the plain single-function $\frac{0}{0}$ up to composed series. Each box is labelled with its example number; that same label heads the worked example below.

A — Approx, small $x > 0$ : plug value into $P_{N}$ , bound error → Ex 1 ( $cos 0.3$ )
B — Approx, small $x < 0$ : negative arg, signs alternate → Ex 2 ( $e^{- 0.5}$ )
C — Approx, edge of convergence: $ln (1 + x)$ near $∣ x ∣ = 1$ , slow → Ex 3 ( $ln 1.5$ )
D — Limit $\frac{0}{0}$ , one function: cancel lowest power → Ex 4
E — Limit $\frac{0}{0}$ , two functions divided: ratio of leaders → Ex 5
F — Limit $\infty - \infty$ recast to $\frac{0}{0}$ : combine first → Ex 6
G — Limit needing a deeper term: leading terms cancel → Ex 7
H — Word problem (real units): small-angle physics → Ex 8
I — Exam twist: substitute a series into a series → Ex 9

Here $P_{N}$ is the partial sum just defined; "plug value into $P_{N}$ " means evaluate that polynomial at your number.

A · Approximate a small positive input

Figure s03 — Ex 1. The red curve is $cos x$ ; the yellow dashed curve is $P_{4} (x) = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24}$ . At the blue marker $x = 0.3$ the two are indistinguishable — the vertical gap (the error) is about $1 0^{- 6}$ , far below our $1 0^{- 4}$ target.

Worked example

Ex 1 (Cell A). Approximate $cos (0.3)$ with error $< 1 0^{- 4}$ . Forecast: guess — will two terms be enough, or three? Write your guess before reading.

Step 1. Use $cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots$ Why this step? $cos$ is one of the four must-own Maclaurin series; near $0$ its terms shrink fast, so a few terms will do.

Step 2. This series alternates (signs $+, -, +, -$ ) and terms decrease, so by the Alternating Series Test the error is $\leq$ the first omitted term. Try keeping through $x^{2}$ : first omitted is $\frac{x ^{4}}{24} = \frac{0. 3 ^{4}}{24} = \frac{0.0081}{24} = 3.4 \times 1 0^{- 4}$ . Too big. Why this step? We test whether two terms clear the $1 0^{- 4}$ bar. They don't — the omitted term is bigger than our target.

Step 3. Keep through $x^{4}$ . First omitted is now $\frac{x ^{6}}{720} = \frac{0. 3 ^{6}}{720} = \frac{0.000729}{720} = 1.0 \times 1 0^{- 6} < 1 0^{- 4}$ . ✓ Why this step? One more term crushes the error far below target.

Step 4. Compute: $1 - \frac{0.09}{2} + \frac{0.0081}{24} = 1 - 0.045 + 0.0003375 = 0.9553375.$ Why this step? Pure arithmetic on the polynomial $P_{4} (0.3)$ .

Verify: true $cos 0.3 = 0.95533649 \dots$ ; our answer $0.9553375$ differs by $\approx 1.0 \times 1 0^{- 6}$ , matching the predicted bound. ✓

B · Approximate a small negative input

Figure s04 — Ex 2. Bars show the size of each successive term $\frac{∣ x ∣ ^{n}}{n !}$ for $x = - 0.5$ ; the colours alternate ( $+$ blue, $-$ red) because odd powers of a negative number flip sign. The bars shrink fast — the first grey (omitted) bar past $x^{4}$ is the error bound.

Worked example

Ex 2 (Cell B). Approximate $e^{- 0.5}$ with error $< 1 0^{- 3}$ . Forecast: with a negative argument the $e^{x}$ terms flip sign. Guess how many terms.

Step 1. $e^{x} = 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \frac{x ^{4}}{24} + \dots$ , put $x = - 0.5$ : $e^{- 0.5} = 1 - 0.5 + \frac{0.25}{2} - \frac{0.125}{6} + \frac{0.0625}{24} - \dots$ Why this step? Odd powers of a negative number are negative, so the series becomes alternating — that gives us the easy alternating-error rule for free.

Step 2. Because it now alternates with shrinking terms, error $\leq$ first omitted term. Keep through $x^{3}$ ; first omitted is $\frac{x ^{4}}{24} = \frac{0.0625}{24} = 2.6 \times 1 0^{- 3}$ . Still too big. Why this step? Same alternating-series logic as Ex 1, applied to check the bound.

Step 3. Keep through $x^{4}$ ; first omitted is $\frac{∣ x ∣ ^{5}}{120} = \frac{0.03125}{120} = 2.6 \times 1 0^{- 4} < 1 0^{- 3}$ . ✓ Why this step? We add the next term to shrink the first-omitted term below the required $1 0^{- 3}$ ; by the Alternating Series Test that omitted term is the error ceiling, so once it drops under target we are guaranteed accurate enough and can stop.

Step 4. Sum through $x^{4}$ : $1 - 0.5 + 0.125 - 0.0208333 + 0.0026042 = 0.6067708.$ Why this step? Arithmetic on the polynomial $P_{4} (- 0.5)$ .

Verify: true $e^{- 0.5} = 0.60653066 \dots$ ; our $0.6067708$ is off by $2.4 \times 1 0^{- 4} < 1 0^{- 3}$ . ✓ (Sanity: $e^{- 0.5} < 1$ since the exponent is negative — good.)

C · Edge of the convergence interval

Figure s05 — Ex 3. The running partial sums of $ln (1.5)$ (blue dots) crawl toward the true value (yellow dashed line) only slowly, ping-ponging above and below it because the series alternates. Contrast the fast $cos$ convergence of Figure s03 — this is what "near the edge" costs you.

Worked example

Ex 3 (Cell C). Estimate $ln (1.5)$ and explain why this is the hard case. Forecast: $ln (1 + x)$ with $x = 0.5$ is well inside $∣ x ∣ < 1$ , but the series converges slowly. Guess: 4 terms, or 8?

Step 1. $ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \frac{x ^{4}}{4} + \dots$ , valid for $- 1 < x \leq 1$ (Radius & Interval of Convergence). Here $x = 0.5$ is legal. Why this step? You must first check $x$ lies in the interval — using this series at $x = 2$ would be the classic blunder from the parent note.

Step 2. It alternates, so error $\leq$ first omitted term $\frac{0. 5 ^{n + 1}}{n + 1}$ . For 6 decimals we'd need many terms because $0. 5^{n + 1}$ shrinks only by half each time. Keep through $x^{4}$ : first omitted $\frac{0. 5 ^{5}}{5} = \frac{0.03125}{5} = 6.25 \times 1 0^{- 3}$ . Why this step? This shows why it's the hard case — near the edge the terms decay slowly, so you need many of them.

Step 3. Sum through $x^{4}$ : $0.5 - \frac{0.25}{2} + \frac{0.125}{3} - \frac{0.0625}{4} = 0.5 - 0.125 + 0.0416667 - 0.015625 = 0.4010417.$ Why this step? Arithmetic; the answer is only accurate to $\sim 6 \times 1 0^{- 3}$ — deliberately mediocre to prove the point.

Verify: true $ln 1.5 = 0.4054651 \dots$ ; our $0.4010417$ is off by $4.4 \times 1 0^{- 3}$ , safely under the $6.25 \times 1 0^{- 3}$ bound. ✓ (Lesson: far from the center, buy accuracy with many terms — or recenter.)

D · A $\frac{0}{0}$ limit, single function

Figure s01 — Ex 4. The red curve is $1 - cos x$ ; the dashed yellow curve is its leading parabola $\frac{x ^{2}}{2}$ . Near $x = 0$ the two overlap almost perfectly, so their ratio approaches $\frac{1}{2}$ . Axes: horizontal $x$ , vertical value; the blue dot marks the origin where both are $0$ .

Worked example

Ex 4 (Cell D). $x \to 0 lim \frac{1 - cos x}{x ^{2}}$ . Forecast: plug $x = 0$ raw and you get $\frac{1 - 1}{0} = \frac{0}{0}$ . Guess the finite answer.

Step 1. $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ , so $1 - cos x = \frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \dots$ Why this step? Replacing $cos$ by its series exposes the smallest surviving power — the source of the $\frac{0}{0}$ .

Step 2. Divide by $x^{2}$ : $\frac{1}{2} - \frac{x ^{2}}{24} + \dots$ Why this step? The $x^{2}$ in numerator and denominator are the same size — they cancel exactly, curing the $\frac{0}{0}$ .

Step 3. Let $x \to 0$ : every remaining term has an $x$ and dies. Limit $= \frac{1}{2}$ . Why this step? Letting $x \to 0$ kills the leftover $x$ -terms, leaving the constant.

Look at the figure: the red curve $1 - cos x$ and the yellow parabola $\frac{x ^{2}}{2}$ hug each other near $0$ — their ratio is exactly what tends to $\frac{1}{2}$ .

Verify: L'Hôpital's Rule twice: $\frac{s i n x}{2 x} \to \frac{c o s x}{2} = \frac{1}{2}$ . Same answer. ✓

E · A $\frac{0}{0}$ limit, two functions divided

Figure s06 — Ex 5. The red curve is $tan x - x$ ; the yellow dashed curve is its leading cubic $\frac{x ^{3}}{3}$ . Both are flat-then-rising near $0$ (a cubic, not a parabola) — dividing by $x^{3}$ leaves the constant $\frac{1}{3}$ , the blue horizontal line the ratio approaches.

F · An $\infty - \infty$ limit recast as $\frac{0}{0}$

Figure s07 — Ex 6. Left: $\frac{1}{x}$ (blue) and $\frac{1}{s i n x}$ (red) both rocket to $\infty$ as $x \to 0$ — subtracting two infinities is meaningless. Right: their difference (green) is a tame curve sliding smoothly through $0$ — that is the value the limit finds.

G · When the leading terms cancel — dig deeper

Figure s08 — Ex 7. A term-by-term ledger of the numerator $e^{x^{2}} - 1 - x^{2}$ : the constant term and the $x^{2}$ term (red, struck through) cancel against the $- 1 - x^{2}$ ; the first survivor is $\frac{x ^{4}}{2}$ (green). Stop before $x^{4}$ and you would wrongly see " $0$ " — this is why you dig deeper.

Worked example

Ex 7 (Cell G). $x \to 0 lim \frac{e ^{x^{2}} - 1 - x ^{2}}{x ^{4} cos x}$ . Forecast: the trap — if you stop too early you'll wrongly get $\frac{0}{0}$ again. Guess.

Step 1. Put $u = x^{2}$ into $e^{u} = 1 + u + \frac{u ^{2}}{2} + \dots$ : $e^{x^{2}} = 1 + x^{2} + \frac{x ^{4}}{2} + \frac{x ^{6}}{6} + \dots$ Why this step? Substituting a series into a series is legal and instant — no chain-rule mess.

Step 2. $e^{x^{2}} - 1 - x^{2} = \frac{x ^{4}}{2} + \frac{x ^{6}}{6} + \dots$ — the $1$ and $x^{2}$ cancel, leaving $x^{4}$ as leader. Why this step? This is the mistake-#1 warning made concrete: you must keep terms up to $x^{4}$ or the numerator "vanishes."

Step 3. Denominator: $x^{4} cos x = x^{4} (1 - \frac{x ^{2}}{2} + \dots) = x^{4} - \frac{x ^{6}}{2} + \dots$ , leader $x^{4}$ . Why this step? Match the denominator's leading degree to the numerator's.

Step 4. Ratio of leaders: $\frac{x ^{4} /2}{x ^{4}} = \frac{1}{2} \to \frac{1}{2}$ . Why this step? Numerator and denominator now share the same leading power $x^{4}$ , so that $x^{4}$ cancels exactly and the surviving quotient is the ratio of the two leading coefficients, $\frac{1/2}{1} = \frac{1}{2}$ ; the remaining terms all carry extra factors of $x$ and vanish as $x \to 0$ .

Verify: at $x = 0.05$ : numerator $= e^{0.0025} - 1 - 0.0025 = 3.1276 \times 1 0^{- 6}$ , denom $= 0.0 5^{4} cos 0.05 = 6.2492 \times 1 0^{- 6}$ , ratio $= 0.5003 \approx \frac{1}{2}$ . ✓

H · Real-world word problem (keep the units)

Figure s09 — Ex 8. A pendulum bob swung out by angle $θ_{0}$ . The vertical rise of the bob above its lowest point is proportional to $1 - cos θ_{0}$ ; the yellow bracket marks that small height, which the series approximates as $\frac{θ _{0}^{2}}{2}$ .

Worked example

Ex 8 (Cell H). For a pendulum, the height the bob rises above its lowest point is proportional to $1 - cos θ_{0}$ , where $θ_{0}$ is the swing angle in radians. For a $θ_{0} = 0.2 rad$ swing, evaluate $1 - cos θ_{0}$ . (Units: $θ_{0}$ in radians, $1 - cos θ_{0}$ is dimensionless.) Forecast: small angle — guess the order of magnitude.

Step 1. $1 - cos θ_{0} \approx \frac{θ _{0}^{2}}{2}$ (from $cos θ = 1 - \frac{θ ^{2}}{2} + \dots$ ). Why this step? Radians are dimensionless, so $θ_{0}^{2}$ is a pure number — the small-angle series is exactly the right tool for physics approximations.

Step 2. $\frac{0. 2 ^{2}}{2} = \frac{0.04}{2} = 0.02$ . Why this step? Arithmetic on the leading term.

Step 3. We drop the next term $\frac{θ _{0}^{4}}{24} = \frac{0.0016}{24} = 6.7 \times 1 0^{- 5}$ because it is a thousand times smaller than $0.02$ — invisible at two significant figures. So $1 - cos (0.2) \approx 0.02$ , i.e. the bob's rise is about a $2%$ effect. Why this step? Keeping only the leading term is justified precisely when the first dropped term is negligible against it; here $6.7 \times 1 0^{- 5} ≪ 0.02$ , so one term is honest, and $0.02$ read as a fraction is $2%$ .

Verify: true $1 - cos 0.2 = 0.0199667$ ; our $0.02$ is off by $3.3 \times 1 0^{- 5}$ , matching the omitted-term size. ✓ (Dimensionless in, dimensionless out — units consistent.)

I · Exam twist — a series inside a series

Figure s10 — Ex 9. The red curve is $ln (1 + sin x) - x$ ; the yellow dashed curve is its leading term $- \frac{x ^{2}}{2}$ . They coincide near $0$ (both open downward), so dividing by $x^{2}$ gives the constant $- \frac{1}{2}$ , the blue horizontal line.

Worked example

Ex 9 (Cell I). $x \to 0 lim \frac{ln ( 1 + sin x ) - x}{x ^{2}}$ . Forecast: two series must be composed. Guess.

Step 1. Inner: $sin x = x - \frac{x ^{3}}{6} + \dots$ . Feed into $ln (1 + u) = u - \frac{u ^{2}}{2} + \dots$ with $u = sin x$ . Why this step? Composition of series is allowed near $0$ because $sin x \to 0$ , keeping us inside $ln (1 + u)$ 's convergence region.

Step 2. Keep to $x^{2}$ : $u = x - \frac{x ^{3}}{6} + \dots, u^{2} = x^{2} + \dots$ $ln (1 + sin x) = (x - \frac{x ^{3}}{6}) - \frac{1}{2} (x^{2}) + \dots = x - \frac{x ^{2}}{2} + \dots$ Why this step? We only need up to $x^{2}$ because the denominator is $x^{2}$ ; higher powers won't survive the crawl.

Step 3. Numerator: $ln (1 + sin x) - x = - \frac{x ^{2}}{2} + \dots$ . Divide by $x^{2}$ : $- \frac{1}{2} + \dots \to - \frac{1}{2}$ . Why this step? The $x$ from $ln (1 + sin x)$ cancels the lone $- x$ , leaving $- \frac{x ^{2}}{2}$ as the leader; dividing by $x^{2}$ matches degrees so the $x^{2}$ cancels, and letting $x \to 0$ kills every remaining $x$ -term, leaving the constant $- \frac{1}{2}$ .

Verify: at $x = 0.01$ : $ln (1 + sin 0.01) - 0.01 = - 5.00 \times 1 0^{- 5}$ , over $x^{2} = 1 0^{- 4}$ gives $- 0.500$ . ✓

Recall Quick self-test (reveal answers)

$lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$ ::: $\frac{1}{2}$ $lim_{x \to 0} \frac{t a n x - x}{x ^{3}}$ ::: $\frac{1}{3}$ $lim_{x \to 0} (\frac{1}{x} - \frac{1}{s i n x})$ ::: $0$ $lim_{x \to 0} \frac{l n ( 1 + s i n x ) - x}{x ^{2}}$ ::: $- \frac{1}{2}$ Which series has interval $- 1 < x \leq 1$ ? ::: $ln (1 + x)$

Connections

Parent: Applications — approximation & limits
Taylor & Maclaurin Series
Lagrange Remainder & Error Estimation
Alternating Series Test
Radius & Interval of Convergence
L'Hôpital's Rule
Power Series — Operations (add, multiply, differentiate)