Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e: Worked examples

The scenario matrix

Every problem built on these two limits falls into one of the cells below. The right column names the example that lands in that cell.

#	Cell (what makes it different)	Which limit	Trap it hides	Example
1	$\frac{s i n ( scaled )}{scaled}$ , plain rescale	sin	forgetting to match arguments	A
2	Higher power $x^{3}$ , needs the $\frac{1 - c o s x}{x ^{2}}$ helper	sin	splitting the power wrong	B
3	Two-sided check: does the limit exist as $x \to 0^{-}$ too? (sign of $x$ )	sin	assuming even/odd blindly	C
4	Geometric / word problem (arc vs chord shrinking)	sin	reading the picture wrong	D
5	Degrees, not radians (the " $= 1$ is a lie" case)	sin	using radian value in degrees	E
6	$(1 + \frac{a}{x})^{x}$ with negative $a$	$e$	sign of the exponent	F
7	Base doesn't shrink like $\frac{1}{x}$ (degenerate $e$ )	$e$	pattern-matching too fast	G
8	Real-world compound growth (continuous interest)	$e$	choosing the wrong $a$	H
9	Exam twist: mix both limits in one expression	both	evaluating pieces separately when you can't	I

Rows 1–5 stress the sin limit across sign, power, dimension, and units. Rows 6–8 stress the $e$ limit across sign and degeneracy. Row 9 is the boss fight.

Before anything, the two engines we keep reusing. Everything below is one of these, disguised:

Example A — plain rescale (cell 1)

Worked example Compute

$x \to 0 lim \frac{sin 7 x}{sin 2 x}$ .

Forecast: guess the answer before reading. (Hint: both shrink like their arguments — so it's a ratio of the arguments.)

Step 1. Multiply and divide to force each sin over its own argument: $\frac{s i n 7 x}{s i n 2 x} = \frac{s i n 7 x}{7 x} \cdot \frac{2 x}{s i n 2 x} \cdot \frac{7 x}{2 x} .$ Why this step? The master $\frac{s i n u}{u}$ needs the argument on top and bottom to match. Here we have two different sines, so we give each its matching denominator, then correct with the leftover $\frac{7 x}{2 x}$ .

Step 2. As $x \to 0$ : $7 x \to 0$ so $\frac{s i n 7 x}{7 x} \to 1$ ; likewise $\frac{2 x}{s i n 2 x} \to 1$ (reciprocal of a $\to 1$ quantity is still $1$ ). The last factor $\frac{7 x}{2 x} = \frac{7}{2}$ is already constant. Why this step? We use the master limit once per sine, then multiply the surviving numbers.

Step 3. $1 \cdot 1 \cdot \frac{7}{2} = \frac{7}{2} .$

Verify: plug a tiny $x = 0.001$ : $sin (0.007) / sin (0.002) \approx 0.006999.../0.001999... \approx 3.4999 \approx \frac{7}{2}$ . ✓

Example B — higher power, needs the cosine helper (cell 2)

Example C — the sign of $x$ : does the limit exist from both sides? (cell 3)

Example D — geometric word problem: arc vs chord (cell 4)

Worked example A point rides on a unit circle. As the central angle

$x$ (radians) shrinks, what is the limiting ratio of the straight chord length to the arc length?

Forecast: chord is a shortcut across the arc, so chord $<$ arc always. But as the angle shrinks, does the ratio go to $0$ , some fraction, or $1$ ?

Step 1. Set up the lengths on the picture.

Arc length on a unit circle $= x$ (this is exactly why Radian Measure is used — the angle is the arc).
Chord length between the two endpoints $= 2 sin \frac{x}{2}$ (drop a perpendicular; each half is $sin \frac{x}{2}$ ). Why this step? We turn a geometry question into a ratio of two familiar quantities.

Step 2. Form the ratio and reshape it to a master limit: $\frac{chord}{arc} = \frac{2 s i n \frac{x}{2}}{x} = \frac{s i n \frac{x}{2}}{\frac{x}{2}} .$ Why this step? Writing $x = 2 \cdot \frac{x}{2}$ makes the argument of sin equal the denominator — the exact shape $\frac{s i n u}{u}$ with $u = \frac{x}{2}$ .

Step 3. As $x \to 0$ , $u = \frac{x}{2} \to 0$ , so the ratio $\to 1$ .

Verify: $x = 0.1$ rad: chord $= 2 sin (0.05) = 0.099979$ , arc $= 0.1$ , ratio $= 0.99979 \to 1$ . ✓ The "thin pizza slice" intuition made numeric.

Example E — degrees, not radians (cell 5)

Worked example Compute

$x \to 0 lim \frac{sin ( x ^{\circ} )}{x}$ , where $x$ is measured in degrees.

Forecast: most people blurt "1". It is not 1. Guess what breaks.

Step 1. Convert degrees to radians inside the sine, because $sin$ in all our theory eats radians: $x^{\circ} = \frac{π}{180} x$ radians. Why this step? The master limit $\frac{s i n u}{u} = 1$ was proven with sector area $\frac{1}{2} r^{2} θ$ , valid only in radians. So we must feed sin a radian argument.

Step 2. Substitute and match arguments: $\frac{s i n ( \frac{π}{180} x )}{x} = \frac{s i n ( \frac{π}{180} x )}{\frac{π}{180} x} \cdot \frac{π}{180} .$ Why this step? Give the sine its matching denominator $\frac{π}{180} x$ ; the correction factor $\frac{π}{180}$ is left over.

Step 3. The matched fraction $\to 1$ , so $lim_{x \to 0} \frac{s i n ( x ^{\circ} )}{x} = \frac{π}{180} \approx 0.01745.$

Verify: $x = 1$ degree: $sin (1^{\circ}) = 0.0174524$ , divided by $1 = 0.0174524 \approx \frac{π}{180}$ . ✓ The clean " $= 1$ " was a radians-only privilege.

Example F — the $e$ limit with a negative constant (cell 6)

Example G — degenerate $e$ : base shrinks too fast (cell 7)

Example H — real-world continuous compounding (cell 8)

Worked example A bank offers

6% annual interest, compounded continuously (infinitely often). By what factor does a deposit grow in 1 year? In 10 years?

Forecast: 6% simple would give $\times 1.06$ in a year. Continuous compounding gives slightly more. Guess how much more.

Step 1. Split the year into $n$ tiny periods; each period pays $\frac{0.06}{n}$ , applied $n$ times: $factor_{1} = (1 + \frac{0.06}{n})^{n} .$ Why this step? "Compounded $n$ times" literally means multiply the base $1 + \frac{rate}{n}$ by itself $n$ times.

Step 2. Let $n \to \infty$ (continuous). This is the master with $a = 0.06$ : $(1 + \frac{0.06}{n})^{n} \to e^{0.06} \approx 1.06184.$ Why this step? Continuous = infinitely many splits = the exact limit that defines $e^{a}$ ; see The Number e and ln.

Step 3. Over $10$ years the exponent scales: factor $= e^{0.06 \times 10} = e^{0.6} \approx 1.82212$ . Why this step? Ten years of continuous growth stacks the rate: $e^{r t}$ with $r = 0.06$ , $t = 10$ .

Verify: $e^{0.06} = 1.061837$ (vs simple $1.06$ — about $0.18%$ more, as forecast). $e^{0.6} = 1.822119$ , so a $1000 deposit becomes about $1822.12. ✓

Example I — the boss fight: both limits at once (cell 9)

Worked example Compute

$x \to \infty lim (cos \frac{1}{x})^{x^{2}}$ .

Forecast: as $x \to \infty$ , $\frac{1}{x} \to 0$ so $cos \frac{1}{x} \to 1$ — another $1^{\infty}$ . Both the cosine limit and the $e$ machinery are hiding here. Guess: does it go to $1$ , $e^{- 1/2}$ , or $0$ ?

Step 1. Log it: $ln (expr) = x^{2} ln (cos \frac{1}{x})$ . Why this step? $1^{\infty}$ — log first, always.

Step 2. Set $t = \frac{1}{x} \to 0$ . Then $x^{2} = \frac{1}{t ^{2}}$ and we need $\frac{l n ( c o s t )}{t ^{2}}$ . Write $ln (cos t) = ln (1 - (1 - cos t)) \approx - (1 - cos t)$ for small $t$ : $\frac{l n ( c o s t )}{t ^{2}} \approx \frac{- ( 1 - c o s t )}{t ^{2}} = - \frac{1 - c o s t}{t ^{2}} \to - \frac{1}{2} .$ Why this step? Here the sin-family helper $\frac{1 - c o s t}{t ^{2}} \to \frac{1}{2}$ (from the parent) plugs directly into the log expansion $ln (1 + u) \approx u$ . Two atomic limits meeting in one line.

Step 3. $ln (expr) \to - \frac{1}{2}$ , so expr $\to e^{- 1/2} = \frac{1}{e} \approx 0.6065$ .

Verify: $x = 1000$ : $(cos 0.001)^{1 0^{6}} = exp (1 0^{6} ln cos 0.001) = exp (1 0^{6} \cdot (- 5.0 \times 1 0^{- 7})) = exp (- 0.5) \approx 0.60653$ . ✓ Not $1$ — the compounding wins.

Recall

Recall Which cell is which — cover and name the trap

Example E's trap ::: using the radian value $1$ when the argument is in degrees; true answer $\frac{π}{180}$ .
Example G's trap ::: pattern-matching to $e$ when the base bump is $\frac{1}{n ^{2}}$ , not $\frac{1}{n}$ ; answer is $1$ .
Example F's trap ::: dropping the sign of $a$ ; answer is $e^{- 4}$ , not $e^{4}$ .
Tool that makes Example B possible ::: the Taylor Series of $sin x$ , since $\frac{s i n x}{x}$ alone can't reach an $x^{3}$ order limit.
Why Example C's limit is even ::: $sin$ is odd, denominator flips too, minus signs cancel.

Connections

Squeeze (Sandwich) Theorem — proves the $\frac{s i n u}{u}$ engine behind A, C, D.
Radian Measure — the pivot of Examples D and E.
Derivative of sin and cos — Example B's slope is the same object seen as a Taylor tail.
The Number e and ln — defines the growth in F, G, H.
Indeterminate Forms — every "log it first" moment.
L'Hôpital's Rule — alternative route for B, F, I (but circular for the atoms themselves).
Taylor Series — the sharpest tool for B and I.