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

Level 1 — Recognition

The whole game at L1: is the argument on top the same as the thing on the bottom? If yes, the ratio is $1$ ; if not, fix it.

Exercise 1.1

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

Recall Solution 1.1

WHAT: the top argument is $7 x$ , the bottom is also $7 x$ . WHY it matters: the atomic limit $\frac{s i n ( stuff )}{same stuff} \to 1$ requires identical arguments — here they already match. Let $u = 7 x$ . As $x \to 0$ , $u \to 0$ , and the expression is exactly $\frac{s i n u}{u} \to 1$ . $1$

Exercise 1.2

Evaluate $x \to 0 lim \frac{tan x}{x}$ .

Recall Solution 1.2

WHAT: split $tan x = \frac{sin x}{cos x}$ . WHY: we want a $\frac{s i n x}{x}$ factor to appear. $\frac{t a n x}{x} = \frac{s i n x}{x} \cdot \frac{1}{c o s x} ⟶ 1 \cdot \frac{1}{1} = 1.$ As $x \to 0$ , $cos x \to 1$ , so the second factor $\to 1$ . $1$

Level 2 — Application

Now top and bottom don't match. Your job: multiply by a clever form of $1$ to force the matching argument, then read off the leftover constant.

Exercise 2.1

Evaluate $x \to 0 lim \frac{sin 4 x}{9 x}$ .

Recall Solution 2.1

WHAT: force $4 x$ under the sin. WHY: the atomic limit needs $\frac{s i n 4 x}{4 x}$ , so we insert $\frac{4 x}{4 x} = 1$ . $\frac{s i n 4 x}{9 x} = \frac{s i n 4 x}{4 x} \cdot \frac{4 x}{9 x} = \to 1 \frac{sin 4 x}{4 x} \cdot \frac{4}{9} .$ $\frac{4}{9}$

Exercise 2.2

Evaluate $x \to 0 lim \frac{sin 3 x}{sin 5 x}$ .

Recall Solution 2.2

WHAT: two sines — build a $\frac{s i n}{arg}$ for each. WHY: divide top and bottom each by $x$ and match arguments. $\frac{s i n 3 x}{s i n 5 x} = \frac{\frac{sin 3 x}{3 x} \cdot 3 x}{\frac{sin 5 x}{5 x} \cdot 5 x} = \frac{\frac{sin 3 x}{3 x}}{\frac{sin 5 x}{5 x}} \cdot \frac{3 x}{5 x} ⟶ \frac{1}{1} \cdot \frac{3}{5} .$ $\frac{3}{5}$

Exercise 2.3

Evaluate $n \to \infty lim (1 + \frac{5}{n})^{n}$ .

Recall Solution 2.3

WHAT: recognise the $e$ -pattern with $a = 5$ . WHY: the master rule from the parent, $(1 + \frac{a}{n})^{n} \to e^{a}$ . Check via log: $ln [(1 + \frac{5}{n})^{n}] = n ln (1 + \frac{5}{n}) = \frac{ln ( 1 + 5/ n )}{1/ n} \to 5$ , using the helper limit $lim_{h \to 0} \frac{l n ( 1 + h )}{h} = 1$ (stated up top, proved in Exercise 4.1) with $h = 5/ n \to 0$ . Since $ln$ of the expression $\to 5$ , the expression itself $\to e^{5}$ . $e^{5}$

Level 3 — Analysis

Here you must split a hard limit into a product of atomic pieces, feeding each the exact power it wants.

Exercise 3.1

Evaluate $x \to 0 lim \frac{1 - cos x}{x sin x}$ .

Recall Solution 3.1

WHAT: we use the helper limit $\frac{1 - c o s x}{x ^{2}} \to \frac{1}{2}$ (listed up top). Rewrite the denominator to expose $x^{2}$ . WHY: split $x sin x = x^{2} \cdot \frac{s i n x}{x}$ , so each factor is a known limit. $\frac{1 - cos x}{x sin x} = \frac{1 - cos x}{x ^{2}} \cdot \frac{x ^{2}}{x sin x} = \frac{1 - cos x}{x ^{2}} \cdot \frac{x}{sin x} .$ As $x \to 0$ : $\frac{1 - cos x}{x ^{2}} \to \frac{1}{2}$ and $\frac{x}{sin x} \to 1$ . $\frac{1}{2}$

Exercise 3.2

Evaluate $x \to 0 lim \frac{x - sin x}{x ^{3}}$ using the Taylor expansion $sin x = x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots$ .

Recall Solution 3.2

WHAT: subtract the series from $x$ . WHY: the leading terms cancel, revealing the $x^{3}$ behaviour. (See Taylor Series.) $x - sin x = x - (x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots) = \frac{x ^{3}}{6} - \frac{x ^{5}}{120} + \dots$ Divide by $x^{3}$ : $\frac{x - s i n x}{x ^{3}} = \frac{1}{6} - \frac{x ^{2}}{120} + \dots ⟶ \frac{1}{6} .$ $\frac{1}{6}$

Exercise 3.3

Evaluate $n \to \infty lim (\frac{2 n + 3}{2 n - 1})^{n}$ .

Recall Solution 3.3

WHAT: rewrite the base as $1 + \frac{something}{2 n - 1}$ . WHY: to match the $e$ -pattern we need $1 + \frac{u}{denominator}$ . $\frac{2 n + 3}{2 n - 1} = \frac{( 2 n - 1 ) + 4}{2 n - 1} = 1 + \frac{4}{2 n - 1} .$ Now match the exponent to the denominator: $(1 + \frac{4}{2 n - 1})^{n} = [(1 + \frac{4}{2 n - 1})^{2 n - 1}]^{\frac{n}{2 n - 1}} .$ Inner bracket $\to e^{4}$ ; exponent $\frac{n}{2 n - 1} \to \frac{1}{2}$ . $\to (e^{4})^{1/2} = e^{2} .$ $e^{2}$

Level 4 — Synthesis

Rebuild the tools and combine both families in one problem.

Exercise 4.1

Prove $h \to 0 lim \frac{ln ( 1 + h )}{h} = 1$ without L'Hôpital, using only the definition $e = lim_{n \to \infty} (1 + \frac{1}{n})^{n}$ and continuity of $ln$ .

Recall Solution 4.1

WHAT: turn the $e$ -limit into a statement about $ln$ . WHY: $ln$ is continuous, so it commutes with limits.

Right side, $h \to 0^{+}$ . Set $h = \frac{1}{n}$ , so $h \to 0^{+}$ corresponds to $n \to \infty$ . Then $\frac{l n ( 1 + h )}{h} = n ln (1 + \frac{1}{n}) = ln (1 + \frac{1}{n})^{n} .$ Because $ln$ is continuous, we may pass the limit inside: $lim_{n \to \infty} ln (1 + \frac{1}{n})^{n} = ln (lim_{n \to \infty} (1 + \frac{1}{n})^{n}) = ln e = 1.$

Left side, $h \to 0^{-}$ . Now $h < 0$ , so $h = \frac{1}{n}$ (positive) can not reach it — we substitute $h = - \frac{1}{n}$ with $n \to \infty$ instead. Then $\frac{l n ( 1 + h )}{h} = \frac{l n ( 1 - \frac{1}{n} )}{- \frac{1}{n}} = n ln (\frac{1}{1 - \frac{1}{n}}) = n ln (\frac{n}{n - 1}) = n ln (1 + \frac{1}{n - 1}) .$ Write $n = (n - 1) + 1$ so the exponent matches the denominator: $ln [(1 + \frac{1}{n - 1})^{n - 1} \cdot (1 + \frac{1}{n - 1})^{1}] \to ln (e \cdot 1) = 1.$ Both one-sided limits equal $1$ , so the two-sided limit is $1$ . $■$ (This is exactly the derivative of $ln x$ at $x = 1$ — see The Number e and ln.)

Exercise 4.2

Evaluate $x \to 0 lim \frac{e ^{2 x} - 1}{sin 3 x}$ . (Use $lim_{u \to 0} \frac{e ^{u} - 1}{u} = 1$ , the helper listed at the top of the page.)

Recall Solution 4.2

WHAT: two atomic limits meet — the exponential ratio and the sine ratio. WHY: build each ratio by inserting matching arguments. $\frac{e ^{2 x} - 1}{s i n 3 x} = \to 1 \frac{e ^{2 x} - 1}{2 x} \cdot \to 1 \frac{3 x}{sin 3 x} \cdot \frac{2 x}{3 x} .$ The last factor is $\frac{2}{3}$ . $\frac{2}{3}$

Level 5 — Mastery

Edge cases, degenerate inputs, and the full geometric proof.

Exercise 5.1 (edge case)

What is $n \to \infty lim (1 + \frac{1}{n ^{2}})^{n}$ ? And $n \to \infty lim (1 + \frac{1}{n})^{n}$ ?

Recall Solution 5.1

Principle: for $(1 + a_{n})^{b_{n}}$ with $a_{n} \to 0$ , log gives $b_{n} ln (1 + a_{n}) \approx b_{n} a_{n}$ (using $\frac{l n ( 1 + h )}{h} \to 1$ ). The limit is $e^{l i m b_{n} a_{n}}$ .

First: $a_{n} = \frac{1}{n ^{2}}$ , $b_{n} = n$ , product $= n \cdot \frac{1}{n ^{2}} = \frac{1}{n} \to 0$ . So limit $= e^{0} = 1$ . $1$ WHY: the base creeps toward $1$ faster than the exponent grows, so the compounding never accumulates — it fizzles to $1$ .

Second: $a_{n} = \frac{1}{n}$ , $b_{n} = n$ , product $= \frac{n}{n} = n \to \infty$ . So limit $= e^{+ \infty} = + \infty$ . $+ \infty$ WHY: now the base grows too slowly to vanish relative to the exponent — the product blows up. Only the exact balance $a_{n} = \frac{c}{n}$ (product $\to c$ ) lands on a finite $e^{c}$ .

Exercise 5.2 (sign / two-sided case)

Is $x \to 0 lim \frac{sin ∣ x ∣}{x}$ defined? Analyse $x \to 0^{+}$ and $x \to 0^{-}$ separately.

Recall Solution 5.2

WHAT: $∣ x ∣$ makes the numerator ignore sign, but the denominator keeps it. WHY split: the two one-sided limits may disagree.

$x \to 0^{+}$ : here $∣ x ∣ = x$ , so $\frac{s i n x}{x} \to 1$ .
$x \to 0^{-}$ : here $∣ x ∣ = - x$ , so $\frac{s i n ( - x )}{x} = \frac{- s i n x}{x} \to - 1$ (using $sin (- x) = - sin x$ and $\frac{s i n x}{x} \to 1$ ).

Left limit $= - 1$ , right limit $= + 1$ . They differ, so: $the two-sided limit does not exist.$ Contrast with the plain $\frac{s i n x}{x}$ , which is even ( $\frac{s i n ( - x )}{- x} = \frac{s i n x}{x}$ ) and so does have limit $1$ from both sides.

Exercise 5.3 (the full geometric proof, all cases)

Prove $lim_{x \to 0} \frac{s i n x}{x} = 1$ for $x \to 0^{+}$ using areas, then justify $x \to 0^{-}$ .

Recall Solution 5.3

Setup (see figure below). On the unit circle take a central angle $x$ with $0 < x < \frac{π}{2}$ . Look at three nested regions.

Figure — Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e

Figure — Unit circle of radius $1$ with central angle $x$ . The white inner triangle (centre, the point $(1, 0)$ , and the circle point) has height $sin x$ , drawn in amber, and area $\frac{s i n x}{2}$ . The cyan shaded wedge is the circular sector, area $\frac{x}{2}$ . The amber outer triangle rises to the tangent line at height $tan x$ (cyan segment), area $\frac{t a n x}{2}$ . Reading outward the shapes strictly nest: white $\subset$ cyan $\subset$ amber, which is exactly the inequality chain below.

Because inner triangle $\subset$ sector $\subset$ outer triangle: $\frac{s i n x}{2} \leq \frac{x}{2} \leq \frac{t a n x}{2} .$ Multiply by $2$ , divide by $sin x > 0$ (direction preserved since positive): $1 \leq \frac{x}{s i n x} \leq \frac{1}{c o s x} .$ Take reciprocals (this flips the inequalities): $cos x \leq \frac{s i n x}{x} \leq 1.$ As $x \to 0^{+}$ , $cos x \to 1$ ; the right bound is $1$ . By the Squeeze (Sandwich) Theorem the middle is trapped, so $\frac{s i n x}{x} \to 1$ . Left side ( $x \to 0^{-}$ ): $\frac{s i n x}{x}$ is even — replacing $x$ by $- x$ leaves it unchanged because both $sin$ and the denominator flip sign together. So the left limit equals the right limit, and the full two-sided limit is $1$ . $■$

Connections

Squeeze (Sandwich) Theorem — the engine for every $\frac{s i n x}{x}$ exercise here.
Radian Measure — why L5 proofs collapse in degrees.
The Number e and ln — L2/L3/L4 $e$ -limits and the $\frac{l n ( 1 + h )}{h}$ tool.
Taylor Series — the shortcut in 3.2.
Indeterminate Forms — $0/0$ and $1^{\infty}$ everywhere on this page.
Derivative of sin and cos — the payoff these limits unlock.
L'Hôpital's Rule — an alternative to 3.2 / 4.2 (but circular for the atomic limits).