Applications — approximation, evaluating limits: Visual walkthrough

Step 1 — The problem: a fraction that refuses to be evaluated

WHAT. We want the value that $\frac{sin x - x}{x ^{3}}$ approaches as $x$ slides toward $0$ .

WHY it is hard. Plug $x = 0$ directly: top is $sin 0 - 0 = 0$ , bottom is $0^{3} = 0$ . We get $\frac{0}{0}$ — a meaningless symbol, not a number. It could be anything. The zero on top and the zero on bottom are racing to zero, and the answer depends entirely on which one gets there faster.

PICTURE. Watch both the top curve $y = sin x - x$ and the bottom curve $y = x^{3}$ dive into the origin. Near $0$ they hug each other — that is the whole mystery in one image.

Figure — Applications — approximation, evaluating limits

Step 2 — The one idea we borrow: every function starts like a polynomial

WHAT. Near $x = 0$ , a smooth function $f$ can be replaced by its Maclaurin series — an infinite polynomial built from $f$ 's value and all its slopes at $0$ : $f (x) = height f (0) + slope piece f^{'} (0) x + bend piece \frac{f ^{''} ( 0 )}{2 !} x^{2} + next piece \frac{f ^{'''} ( 0 )}{3 !} x^{3} + \dots$ Each term is labelled by what it does: the constant sets the height, the $x$ -term sets the tilt, the $x^{2}$ -term sets the curvature, and so on.

WHY this tool and not another. We could use L'Hôpital's Rule — differentiate top and bottom repeatedly. But that hides why the answer is what it is. The series lays every "speed of approach" out in plain sight: the lowest power of $x$ is exactly how fast that function leaves zero. That is precisely the information the tug-of-war needs.

PICTURE. Adding one more term at a time makes the polynomial hug $sin x$ over a wider stretch. The dashed accent curve creeps outward toward the black true curve.

Recall Why

$\frac{f ^{(n)} ( 0 )}{n !}$ ? (the coefficient, in one line) Differentiating $x^{n}$ exactly $n$ times leaves $n!$ ; every other term dies at $x = 0$ . So to recover $f^{(n)} (0)$ the coefficient must carry a $\frac{1}{n !}$ . Full derivation lives in Taylor & Maclaurin Series.

Step 3 — Read off the "leaving-zero speed" of the numerator

WHAT. Expand the top, $sin x - x$ , using the standard series $sin x = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots$ : $sin x - x = (cancels x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots) - cancels x = - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots$ Here $\frac{x ^{3}}{6}$ is $\frac{x ^{3}}{3 !}$ (since $3! = 6$ ), and $\frac{x ^{5}}{120} = \frac{x ^{5}}{5 !}$ .

WHY the $x$ cancels. $sin x$ and $x$ have the same height (0) and same slope (1) at the origin — they leave zero identically at first order. The subtraction wipes out that shared part, exposing the first place they differ: the $x^{3}$ term. So the numerator leaves zero like $x^{3}$ , at "cubic speed."

PICTURE. The straight line $y = x$ and the curve $y = sin x$ are tangent at $0$ ; their gap $sin x - x$ opens up only as a gentle $x^{3}$ dip.

Step 4 — Read off the "leaving-zero speed" of the denominator

WHAT. The bottom is already a pure polynomial: $x^{3}$ . Its leading (and only) power is $x^{3}$ .

WHY this is the easy side. No expansion needed — $x^{3}$ is its own series. Its leaving-zero speed is exactly cubic. Notice the numerator (Step 3) is also cubic. That equality is the punchline coming: a cubic racing a cubic gives a finite, non-zero answer.

PICTURE. Three curves stacked: $x$ (leaves zero slowest, linear), $x^{2}$ (faster), $x^{3}$ (fastest of the three). The accent $x^{3}$ is the one that matches our numerator's degree.

Step 5 — Cancel the shared power and let $x$ crawl to zero

WHAT. Divide the numerator series by $x^{3}$ : $\frac{s i n x - x}{x ^{3}} = \frac{- \frac{1}{6} x ^{3} + \frac{1}{120} x ^{5} - \dots}{x ^{3}} = survives - \frac{1}{6} + \to 0 \frac{1}{120} x^{2} - \dots$ Every term had at least $x^{3}$ , so dividing by $x^{3}$ leaves a constant $- \frac{1}{6}$ plus terms still carrying an $x$ .

WHY this finishes it. Now there is no $\frac{0}{0}$ — the fraction became an ordinary polynomial in $x$ . Send $x \to 0$ : the constant $- \frac{1}{6}$ stays, everything with an $x$ dies. Limit read off directly.

PICTURE. The messy quotient function flattens as $x \to 0$ onto the horizontal accent line at height $- \frac{1}{6}$ .

Step 6 — Edge case A: keeping too FEW terms (the classic trap)

WHAT. Suppose you lazily used $sin x \approx x$ (one term). Then $sin x - x \approx 0$ , and you'd conclude the limit is $\frac{0}{x ^{3}} = 0$ . Wrong.

WHY it fails. You threw away the very term ( $- \frac{x ^{3}}{6}$ ) that survives the division. You must expand the numerator to at least the degree of the denominator — here degree $3$ — or you delete the answer.

PICTURE. A "term budget" bar: one term is not enough, the truth first appears at the $x^{3}$ term (accent), everything beyond is safe to drop.

Step 7 — Edge case B: a tie with a twist ( $x - tan x$ )

WHAT. The parent's trickiest example: $x \to 0 lim \frac{x - sin x}{x - tan x}$ . Expand both: $x - sin x = \frac{x ^{3}}{6} - \dots, x - tan x = x - (x + \frac{x ^{3}}{3} + \dots) = - \frac{x ^{3}}{3} - \dots$ Here $tan x = x + \frac{x ^{3}}{3} + \dots$ , so subtracting from $x$ leaves $- \frac{x ^{3}}{3}$ .

WHY both are cubic. $sin x$ curves below $x$ (dips down $\Rightarrow$ positive $\frac{x ^{3}}{6}$ after the minus flips), $tan x$ curves above $x$ (leading $- \frac{x ^{3}}{3}$ ). Both leave zero at cubic speed but with opposite signs — that sign is what makes the final answer negative. Ratio of leading coefficients: $\frac{x ^{3} /6}{- x ^{3} /3} = \frac{1/6}{- 1/3} = - \frac{1}{2} .$

PICTURE. Around $0$ : $tan x$ rides above the line $y = x$ , $sin x$ rides below it. The two gaps (accent) point in opposite directions — that opposition is the minus sign in the answer.

Step 8 — Edge case C: when the race is NOT a tie

WHAT. Change the denominator to see the other two outcomes:

$x \to 0 lim \frac{sin x - x}{x ^{2}} = x \to 0 lim \frac{- x ^{3} /6 + \dots}{x ^{2}} = x \to 0 lim (- \frac{x}{6} + \dots) = 0.$ Numerator (cubic) beats denominator (quadratic) $\Rightarrow$ collapses to $0$ .
$x \to 0 lim \frac{sin x - x}{x ^{4}} = x \to 0 lim \frac{- x ^{3} /6 + \dots}{x ^{4}} = x \to 0 lim (- \frac{1}{6 x} + \dots) = \mp \infty.$ Denominator (quartic) wins $\Rightarrow$ blows up.

WHY show these. So you never guess. The degree comparison decides before any arithmetic whether you'll get $0$ , a number, or infinity. Only degree-match ( $m = k$ ) gives a finite non-zero value.

PICTURE. Three panels of the numerator's cubic against denominators $x^{2}, x^{3}, x^{4}$ : the accent column marks which one ties.

Recall Sign of the blow-up (

$x^{4}$ denominator) As $x \to 0^{+}$ , $x^{4} > 0$ , so $- \frac{1}{6 x ^{4}} \cdot x^{3} = - \frac{1}{6 x} \to - \infty$ . As $x \to 0^{-}$ , it $\to + \infty$ . The two-sided limit does not exist — worth stating, since a bare " $\infty$ " hides that the sides disagree.

The one-picture summary

WHAT. Everything compressed: numerator and denominator are each ranked by their lowest surviving power of $x$ ; the limit is decided by comparing those two ranks, and equals the ratio of leading coefficients when they match.

Recall Feynman retelling — the whole walkthrough in plain words

A $\frac{0}{0}$ fraction is a race: top and bottom both sprint toward zero, and we ask who is faster. To measure the speeds we swap each curve for its polynomial twin — the twin's smallest power of $x$ tells you how fast it leaves zero. For $sin x - x$ the linear parts cancel (sine and the line $x$ leave zero identically at first), so the top only differs at the $x^{3}$ term: cubic speed. The bottom $x^{3}$ is also cubic. Same speed = a tie, and a tie always gives a clean number — just the ratio of the leading coefficients, $- \frac{1}{6} \div 1 = - \frac{1}{6}$ . If the top had been faster we'd get $0$ ; if the bottom faster, infinity. The only way to lose is to keep too few terms and accidentally erase the term that decides the race — so always carry each series one power past the denominator, then cancel and let $x$ crawl to zero.

Connections

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

Visual walkthrough — Applications — approximation, evaluating limits