Substituting into gives , and no factoring can cancel a sine against an . How can a limit be found when algebra has nothing to grab, and what does sine do near zero?

Calculus 1

The squeeze theorem and the limit of sine x over x

Substituting x = 0 into \dfrac{\sin x}{x} gives \dfrac{0}{0}, and no factoring can cancel a sine against an x. How can a limit be found when algebra has nothing to grab, and what does sine do near zero?

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The Pythagorean identitiesPrecalculus The limit lawsCalculus 1

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Convergence of sequencesCalculus 2 Derivatives of the trigonometric functionsCalculus 1

Where you are 11 in Limits

Calculus8

Limits11

▴ 6 above Continuity5 Infinite limits and vertical asymptotes Limits at infinity and horizontal asymptotes The intermediate value theorem

The squeeze theorem and the limit of sine x over x

Why this exists

The limit laws find a limit from the limits of its pieces, and Limits by algebra: factoring and rationalizing rescues a quotient that comes out as \displaystyle \frac{0}{0} by canceling a common factor. Both fail on two kinds of function that calculus cannot do without. The first wobbles: \displaystyle \cos \frac{2}{x} swings between -1 and 1 infinitely often near 0 and has no limit there, so the product law cannot be applied to \displaystyle x^4 \cos \frac{2}{x}. The second mixes a sine with a power of x: \displaystyle \frac{\sin x}{x} is \displaystyle \frac{0}{0} at 0, and there is no common factor to cancel, because \sin x is not a polynomial.

The squeeze theorem handles both. Instead of computing a limit directly, it traps the function between two simpler functions heading to the same place. Day 1 of this lesson states and proves the theorem and uses it. Day 2 uses it, with a picture of the unit circle, to prove the central limit of trigonometric calculus:

\lim_{x \to 0} \frac{\sin x}{x} = 1

That one fact, with x in radians, is what makes the derivatives of sine and cosine possible.

Definition

The squeeze theorem, in plain words: if a function is trapped between two others near a, and the two outer functions approach the same number L, the trapped one has nowhere else to go, so it approaches L too.

In symbols: suppose that for every x \ne a in some open interval around a,

g(x) \le f(x) \le h(x)

and that

\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L

Then \lim_{x \to a} f(x) = L. Nothing is assumed about the value at a: the functions need not be defined there. The same statement holds for the one-sided limits x \to a^- and x \to a^+, with the inequality needed on that side only, and for x \to \infty, with the inequality needed for every x beyond some number.

The limit of sine over its input. When x is measured in radians,

\lim_{x \to 0} \frac{\sin x}{x} = 1

In plain words: for small angles, \sin x and x are nearly equal, and their ratio can be brought as close to 1 as you like by taking x close enough to 0. Radian measure is part of the statement, not a convenience; in degrees the limit is a different number.

Intuition

Picture the graphs of g and h as the two walls of a funnel narrowing to the single point (a, L) (Figure 1). The graph of f lies between the walls, however it wiggles. Close to a the gap between the walls is tiny, so every output of f there is within that tiny gap of L. The function f may oscillate, may be undefined at a, may have a formula nobody can simplify; none of that matters once it is trapped.

Figure 1The graph of f is trapped between the graphs of g and h, which both approach L as x \to a. The gap closes, so f(x) approaches L too, whatever happens at a itself.

For \displaystyle \frac{\sin x}{x} the walls come from the unit circle (Figure 2). For a small positive angle x, three lengths nearly agree: the height \sin x of the point P, the arc of length x from A to P, and the height \tan x where the ray OP meets the vertical line through A. The arc is longer than \sin x and shorter than \tan x, and as x shrinks the three crowd together, so their ratios approach 1.

Figure 2Triangle OAP sits inside sector OAP, which sits inside triangle OAT. Their areas are \displaystyle \frac{1}{2}\sin x, \displaystyle \frac{1}{2}x and \displaystyle \frac{1}{2}\tan x.

The graph of \displaystyle y = \frac{\sin x}{x} (Figure 3) shows the result: a hole at (0, 1), symmetry about the y axis because \displaystyle \frac{\sin(-x)}{-x} = \frac{\sin x}{x}, a zero at every nonzero multiple of \pi, and waves that die away as |x| grows.

Figure 3The graph of \displaystyle y = \frac{\sin x}{x} has a hole at (0, 1). The function is undefined at 0, but its limit there is 1.

Results

Day 1: the squeeze theorem

The squeeze theorem is the statement in the definition.

A bounded factor times a factor going to zero. If |u(x)| \le M for every x \ne a near a, for some constant M, and \lim_{x \to a} v(x) = 0, then

\lim_{x \to a} u(x)\,v(x) = 0

This holds even when u has no limit at a, which is exactly when the product law cannot be used.

Day 2: sine over its input

With x in radians,

\begin{aligned} \lim_{x \to 0} \frac{\sin x}{x} &= 1 \\ \lim_{x \to 0} \frac{1 - \cos x}{x} &= 0 \\ \lim_{x \to 0} \frac{\tan x}{x} &= 1 \end{aligned}

Rescaling the input. For any constant k \ne 0, \displaystyle \lim_{x \to 0} \frac{\sin kx}{kx} = 1, and the same holds for the other two limits with kx in place of x. What matters is that the input of the sine and the denominator are the same quantity, and that this quantity goes to 0, without equaling 0, as x \to 0.

Sine and cosine at 0. On the way, the proof shows \lim_{x \to 0} \sin x = 0 and \lim_{x \to 0} \cos x = 1.

Degrees. If x is measured in degrees, written x^\circ, then

\lim_{x \to 0} \frac{\sin(x^\circ)}{x} = \frac{\pi}{180} \approx 0.01745

Why it is true

The squeeze theorem. Name a tolerance t > 0. Because g(x) \to L, there is a window around a in which every x \ne a has g(x) > L - t. Because h(x) \to L, there is a window in which h(x) < L + t. And there is a window in which g(x) \le f(x) \le h(x). In the smallest of the three windows, every x \ne a has

L - t < g(x) \le f(x) \le h(x) < L + t

so f(x) is within t of L. Since t was any tolerance, f(x) \to L. The exact meaning of “window” and “tolerance” is in The epsilon delta definition of a limit.

A bounded factor times a factor going to zero. Since |u(x)\,v(x)| = |u(x)|\,|v(x)| \le M|v(x)|, we have -M|v(x)| \le u(x)\,v(x) \le M|v(x)|. As v(x) \to 0, also |v(x)| \to 0, since |v(x)| is exactly the distance from v(x) to 0. The constant multiple law sends both walls to 0, and the squeeze theorem does the rest.

The inequality from the unit circle. Take \displaystyle 0 < x < \frac{\pi}{2}. In Figure 2, O is the center of the unit circle, A = (1, 0), P = (\cos x, \sin x) is the point at angle x, and T = (1, \tan x) is where ray OP meets the vertical line through A. Triangle OAP lies inside sector OAP, which lies inside triangle OAT.

Region	Area	Reason
triangle OAP	\displaystyle \frac{1}{2}\sin x	base OA = 1, height \sin x
sector OAP	\displaystyle \frac{1}{2}x	\displaystyle \frac{1}{2}r^2\theta with r = 1 and \theta = x
triangle OAT	\displaystyle \frac{1}{2}\tan x	legs OA = 1 and AT = \tan x

The sector formula is from Sectors, segments and their areas, and it holds only for an angle in radians (Radian measure). Each region is strictly inside the next, so doubling the areas gives

\sin x < x < \tan x \qquad \text{for } 0 < x < \tfrac{\pi}{2}

Sine and cosine near 0. The left half says 0 < \sin x < x for small positive x. Since \sin(-x) = -\sin x, this gives -|x| \le \sin x \le |x| for \displaystyle |x| < \frac{\pi}{2}, and the squeeze theorem gives \sin x \to 0. For \displaystyle |x| < \frac{\pi}{2} cosine is positive, so the Pythagorean identity \sin^2 x + \cos^2 x = 1 (The Pythagorean identities) gives \displaystyle \cos x = \sqrt{1 - \sin^2 x}, and the power, difference and root laws give \displaystyle \cos x \to \sqrt{1 - 0} = 1.

Sine over its input. For \displaystyle 0 < x < \frac{\pi}{2}, divide \sin x < x < \tan x by the positive number \sin x, using \displaystyle \tan x = \frac{\sin x}{\cos x}:

1 < \frac{x}{\sin x} < \frac{1}{\cos x}

All three are positive, so taking reciprocals reverses both inequalities:

\cos x < \frac{\sin x}{x} < 1

For \displaystyle -\frac{\pi}{2} < x < 0 nothing changes, since \displaystyle \frac{\sin(-x)}{-x} = \frac{\sin x}{x} and \cos(-x) = \cos x. So the inequality holds whenever \displaystyle 0 < |x| < \frac{\pi}{2}. The lower wall \cos x tends to 1 and the upper wall is the constant 1, so the squeeze theorem gives \displaystyle \frac{\sin x}{x} \to 1.

One minus cosine. For \displaystyle 0 < |x| < \frac{\pi}{2}, 1 + \cos x > 0, so multiply top and bottom by it and use 1 - \cos^2 x = \sin^2 x:

\begin{aligned} \frac{1 - \cos x}{x} &= \frac{1 - \cos^2 x}{x(1 + \cos x)} \\ &= \frac{\sin x}{x} \cdot \frac{\sin x}{1 + \cos x} \end{aligned}

By the product and quotient laws, the limit is \displaystyle 1 \cdot \frac{0}{1 + 1} = 0.

Tangent over its input. \displaystyle \frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}, which tends to 1 \cdot 1 = 1.

Rescaling. Let k \ne 0 and u = kx. As x \to 0 with x \ne 0, u \to 0 with u \ne 0, so every value \displaystyle \frac{\sin kx}{kx} is a value of \displaystyle \frac{\sin u}{u} at an input near 0. If \displaystyle \frac{\sin u}{u} is within a tolerance of 1 whenever 0 < |u| < s, then \displaystyle \frac{\sin kx}{kx} is within it whenever \displaystyle 0 < |x| < \frac{s}{|k|}.

Degrees. An angle of x degrees is \displaystyle \frac{\pi x}{180} radians, so \sin(x^\circ) = \sin u with \displaystyle u = \frac{\pi x}{180}, and

\frac{\sin(x^\circ)}{x} = \frac{\pi}{180} \cdot \frac{\sin u}{u} \to \frac{\pi}{180}

In the picture, the sector of x degrees has area \displaystyle \frac{\pi x}{360}, not \displaystyle \frac{x}{2}, and that factor \displaystyle \frac{\pi}{180} rides through every step. Radians are the one unit in which the sector’s area is half the angle, which is why the limit is exactly 1 there and nowhere else.

Worked examples

Example 1

Day 1: Bachisio’s wobbling product

Bachisio wants \lim_{x \to 0} x^4 \cos \dfrac{2}{x}.

Solution

The product law needs both factors to have limits, and \displaystyle \cos \frac{2}{x} has none: as x \to 0 the input \displaystyle \frac{2}{x} grows without bound, so the cosine passes through 1 and -1 over and over. Squeeze instead. Cosine of any input lies between -1 and 1:

-1 \le \cos \frac{2}{x} \le 1 \qquad (x \ne 0)

Multiply through by x^4, which is positive for x \ne 0, so the inequalities keep their direction:

-x^4 \le x^4 \cos \frac{2}{x} \le x^4

The walls are polynomials, so \lim_{x \to 0} (-x^4) = 0 and \lim_{x \to 0} x^4 = 0. By the squeeze theorem,

\lim_{x \to 0} x^4 \cos \frac{2}{x} = 0

The wobble never stops, but its size x^4 shrinks to nothing. This is the bounded-times-zero result with \displaystyle u(x) = \cos \frac{2}{x}, M = 1 and v(x) = x^4.

Example 2

Day 1: Grazia’s function known only by its bounds

Grazia knows nothing about a function f except that 2x \le f(x) \le x^2 + 1 for every x. Find \lim_{x \to 1} f(x). What can be said about \lim_{x \to 0} f(x)?

Solution

First, the bounds are consistent: x^2 + 1 - 2x = (x - 1)^2 \ge 0, so the lower wall never rises above the upper one, and they touch only at x = 1. Both walls are polynomials, so

\lim_{x \to 1} 2x = 2, \qquad \lim_{x \to 1} (x^2 + 1) = 2

The two limits agree, and the squeeze theorem gives \lim_{x \to 1} f(x) = 2.

At 0 the walls tend to 0 and 1, which differ, so the theorem says nothing. In fact the bounds cannot decide: f(x) = 2x and f(x) = x^2 + 1 both satisfy them, and their limits at 0 are 0 and 1. A squeeze works only where the walls meet.

Example 3

Day 1: Efisio’s squeeze at infinity

Efisio wants \lim_{x \to \infty} \dfrac{3x - \cos x}{x}.

Solution

The quotient law is no use, since top and bottom both grow without bound. Split the fraction instead, for x > 0:

\frac{3x - \cos x}{x} = 3 - \frac{\cos x}{x}

Now squeeze the piece \displaystyle \frac{\cos x}{x}. For x > 0, dividing -1 \le \cos x \le 1 by the positive number x gives

-\frac{1}{x} \le \frac{\cos x}{x} \le \frac{1}{x}

As x grows, \displaystyle \frac{1}{x} shrinks toward 0, and so does \displaystyle -\frac{1}{x}. The squeeze theorem at infinity gives \displaystyle \frac{\cos x}{x} \to 0, and the difference law gives

\lim_{x \to \infty} \frac{3x - \cos x}{x} = 3 - 0 = 3

The cosine term is a wobble of fixed size divided by an ever larger number, so it fades out.

Example 4

Day 2: rescaling the input

Find \lim_{x \to 0} \dfrac{\sin 5x}{3x}. What would a calculator set to degrees suggest instead?

Solution

The result \displaystyle \frac{\sin u}{u} \to 1 needs the denominator to match the input of the sine. Here the input is 5x and the denominator is 3x, so supply the 5 and compensate:

\frac{\sin 5x}{3x} = \frac{5}{3} \cdot \frac{\sin 5x}{5x}

As x \to 0, 5x \to 0, so by rescaling \displaystyle \frac{\sin 5x}{5x} \to 1, and by the constant multiple law

\lim_{x \to 0} \frac{\sin 5x}{3x} = \frac{5}{3}

A calculator in degree mode reads the input 5x as degrees, which is \displaystyle \frac{5\pi x}{180} radians. The same rewriting then gives \displaystyle \frac{5}{3} \cdot \frac{\pi}{180} = \frac{\pi}{108} \approx 0.0291. Its table would settle near 0.0291, a correct answer to a different question.

Example 5

Day 2: one minus cosine over the square

Find \lim_{x \to 0} \dfrac{1 - \cos x}{x^2} exactly. (Exercise 7 of What a limit is: graphs and tables estimated it from a table.)

Solution

Substituting gives \displaystyle \frac{0}{0}. Multiply top and bottom by 1 + \cos x, which is positive near 0, and use the Pythagorean identity:

\begin{aligned} \frac{1 - \cos x}{x^2} &= \frac{1 - \cos^2 x}{x^2 (1 + \cos x)} \\ &= \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1 + \cos x} \end{aligned}

By the power law \displaystyle \left(\frac{\sin x}{x}\right)^2 \to 1^2 = 1, and by the quotient law \displaystyle \frac{1}{1 + \cos x} \to \frac{1}{2}. So

\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

Next to the Results, this says that 1 - \cos x behaves like \displaystyle \frac{1}{2}x^2 for small x: so small compared with x that dividing by x still leaves something going to 0, but of the same size as x^2.

Example 6

Day 2: tangent over sine

Bachisio wants \lim_{x \to 0} \dfrac{\tan 3x}{\sin 2x}.

Solution

There are two different inputs, 3x and 2x. Give each its own matching denominator, and put back what was borrowed:

\frac{\tan 3x}{\sin 2x} = \frac{\tan 3x}{3x} \cdot \frac{2x}{\sin 2x} \cdot \frac{3x}{2x}

For x \ne 0 the last factor is \displaystyle \frac{3}{2}. By rescaling, \displaystyle \frac{\tan 3x}{3x} \to 1 and \displaystyle \frac{\sin 2x}{2x} \to 1, so its reciprocal \displaystyle \frac{2x}{\sin 2x} \to 1 by the quotient law. The product law gives

\lim_{x \to 0} \frac{\tan 3x}{\sin 2x} = 1 \cdot 1 \cdot \frac{3}{2} = \frac{3}{2}

Example 7

Day 2: a chord against its arc

Efisio runs a bend of a track that is an arc of a circle of radius 40 meters with central angle \theta radians. Compare the arc he runs with the straight chord joining its ends when \theta = 1, and find the limit of chord over arc as \theta \to 0^+.

Solution

The arc has length 40\theta meters (Arc length and sector area in radians). For the chord, the segment from the center to the chord’s midpoint is perpendicular to the chord and splits the isosceles triangle into two right triangles, each with angle \displaystyle \frac{\theta}{2} at the center and hypotenuse 40. Half the chord is \displaystyle 40 \sin \frac{\theta}{2}, so the chord is \displaystyle 80 \sin \frac{\theta}{2} meters. For \theta = 1 the arc is 40 meters and the chord is 80 \sin 0.5 \approx 38.35 meters.

The ratio is

\frac{\text{chord}}{\text{arc}} = \frac{80 \sin \frac{\theta}{2}}{40\theta} = \frac{\sin \frac{\theta}{2}}{\frac{\theta}{2}}

As \theta \to 0^+, \displaystyle \frac{\theta}{2} \to 0^+, so by rescaling the ratio tends to 1. The radius has canceled: on any circle, a short chord and its arc are nearly the same length, and their ratio can be brought as close to 1 as wanted by shrinking the angle. That is what lets a curve be measured by short straight pieces.

Boundaries

The walls must meet. The squeeze theorem needs \lim g = \lim h. In Example 2 the walls tend to 0 and 1 at x = 0, and the limit of f there is not determined.
The trap must hold on both sides of a for a two-sided limit, at every x \ne a in some open interval around a. A trap on one side only gives a one-sided limit. Far from a, and at a itself, the inequality may fail without harm.
A bound alone traps nothing. \displaystyle |\cos \frac{2}{x}| \le 1, yet \displaystyle \cos \frac{2}{x} has no limit at 0: its walls -1 and 1 do not meet. The bounded factor needs a partner that goes to 0.
Radians only. \displaystyle \frac{\sin x}{x} \to 1 relies on the sector area \displaystyle \frac{1}{2}x, true only in radians. In degrees the limit is \displaystyle \frac{\pi}{180}.
Only as the input goes to 0. At any other point \displaystyle \frac{\sin x}{x} is a quotient with nonzero bottom, and direct substitution applies: as \displaystyle x \to \frac{\pi}{2} it tends to \displaystyle \frac{2}{\pi}. As x \to \infty the limit is something else entirely.
The input and the denominator must match. \displaystyle \frac{\sin 5x}{x} tends to 5, not 1, and \displaystyle \frac{\sin x}{5x} tends to \displaystyle \frac{1}{5}.

Where this goes next

The two limits \displaystyle \frac{\sin x}{x} \to 1 and \displaystyle \frac{1 - \cos x}{x} \to 0 are exactly what the derivative of sine needs: expanding \sin(x + h) by Sine and tangent of a sum or difference turns the difference quotient into these two limits, which gives the derivatives of the trigonometric functions. The proof also showed that sine and cosine approach their values at 0, the first step toward their continuity everywhere. The squeeze theorem works word for word for sequences, where it is one of the main tools of Convergence of sequences. Later, L’Hopital’s rule finds many \displaystyle \frac{0}{0} limits quickly, but it cannot be used to prove \displaystyle \frac{\sin x}{x} \to 1: the rule needs the derivative of sine, and finding that derivative uses this very limit.

Exercises

Day 1: squeezing to a limit

Find \lim_{x \to 0} x^2 \sin \dfrac{5}{x}.
Find \lim_{x \to 0} |x| \cos \dfrac{1}{x^2}.
Find \displaystyle \lim_{x \to 0^+} \sqrt{x}\,\Big(2 + \sin \dfrac{1}{x}\Big).
For every x > 0, 3x - 2 \le f(x) \le x^3. Find \lim_{x \to 1} f(x).
For 0 < |x| < 2, \displaystyle \sqrt{4 - x^2} \le f(x) \le 2 + x^2. Find \lim_{x \to 0} f(x).
Find \displaystyle \lim_{x \to \infty} \dfrac{2 - \sin x}{\sqrt{x}}.
Find \lim_{x \to \infty} \dfrac{4x + \sin(x^2)}{2x - 1}.

Day 2: sine over its input

Find \lim_{x \to 0} \dfrac{\sin 4x}{x}.
Find \lim_{x \to 0} \dfrac{\sin 2x}{\sin 7x}.
Find \lim_{x \to 0} \dfrac{x}{\tan 6x}.
Find \lim_{x \to 0} \dfrac{1 - \cos 3x}{x}.
Find \lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}.
Find \lim_{x \to 0} \dfrac{x \sin x}{1 - \cos x}.
Find \lim_{x \to \pi} \dfrac{\sin x}{\pi - x}.
Find \lim_{x \to 0} \dfrac{\sin(x^2)}{x}.

Day 2: combining the limits

Find \lim_{x \to 0} \dfrac{\tan x - \sin x}{x^3}.
Find \lim_{x \to 0} \dfrac{\cos x - \cos 3x}{x^2}.
With the input of tangent read in degrees, find \lim_{x \to 0} \dfrac{\tan(x^\circ)}{x}.

Reasoning

Efisio writes \lim_{x \to 0} \dfrac{\sin x}{x} = \dfrac{\lim_{x \to 0} \sin x}{\lim_{x \to 0} x} = \dfrac{0}{0}, “so the limit does not exist”. What is wrong?
All that is known about f is that 1 - |x| \le f(x) \le 2 + |x| for every x. What can be concluded about \lim_{x \to 0} f(x)? Give examples.
Grazia knows only that 1 - x^2 \le f(x) \le 1 + x^2 for 0 < |x| < 0.01, and nothing about f anywhere else, not even f(0). Explain why that is enough to find \lim_{x \to 0} f(x), and find it.
Explain why \lim_{x \to \infty} \dfrac{\sin x}{x} is 0 and not 1.
Show that \lim_{x \to 0} \dfrac{\sin kx}{x} = k for every constant k, and explain why the substitution u = kx is allowed when k \ne 0.
Use the areas in Figure 2, with the angle measured as x degrees, to explain why \dfrac{\sin(x^\circ)}{x} tends to \dfrac{\pi}{180} rather than 1.

Answers

0. For x \ne 0, \displaystyle -x^2 \le x^2 \sin \frac{5}{x} \le x^2, and both walls tend to 0.
0. For x \ne 0, \displaystyle -|x| \le |x| \cos \frac{1}{x^2} \le |x|, and |x| \to 0.
0. Since \displaystyle 1 \le 2 + \sin \frac{1}{x} \le 3 and \displaystyle \sqrt{x} > 0 for x > 0, \displaystyle \sqrt{x} \le f(x) \le 3\sqrt{x}. Both walls tend to 0 as x \to 0^+, so the one-sided squeeze gives 0.
1. The walls tend to 3 \cdot 1 - 2 = 1 and 1^3 = 1. (The bounds are consistent: x^3 - 3x + 2 = (x - 1)^2(x + 2) \ge 0 for x > 0.)
2. By the root law \displaystyle \sqrt{4 - x^2} \to \sqrt{4} = 2, and 2 + x^2 \to 2.
0. For x > 0, 1 \le 2 - \sin x \le 3, so \displaystyle \frac{1}{\sqrt{x}} \le \frac{2 - \sin x}{\sqrt{x}} \le \frac{3}{\sqrt{x}}, and both walls tend to 0.
2. Divide top and bottom by x: \displaystyle \frac{4 + \sin(x^2)/x}{2 - 1/x}. Since \displaystyle -\frac{1}{x} \le \frac{\sin(x^2)}{x} \le \frac{1}{x}, that piece tends to 0, and the quotient law gives \displaystyle \frac{4}{2} = 2.
4. \displaystyle \frac{\sin 4x}{x} = 4 \cdot \frac{\sin 4x}{4x} \to 4 \cdot 1.
\displaystyle \frac{2}{7}. \displaystyle \frac{\sin 2x}{2x} \cdot \frac{7x}{\sin 7x} \cdot \frac{2}{7} \to 1 \cdot 1 \cdot \frac{2}{7}.
\displaystyle \frac{1}{6}. \displaystyle \frac{x}{\tan 6x} = \frac{1}{6} \cdot \frac{6x}{\tan 6x}, and \displaystyle \frac{6x}{\tan 6x} \to 1 as the reciprocal of \displaystyle \frac{\tan 6x}{6x} \to 1.
0. \displaystyle \frac{1 - \cos 3x}{x} = 3 \cdot \frac{1 - \cos 3x}{3x} \to 3 \cdot 0.
2. \displaystyle \frac{1 - \cos 2x}{x^2} = 4 \cdot \frac{1 - \cos 2x}{(2x)^2} \to 4 \cdot \frac{1}{2}, using Example 5 rescaled.
2. Divide top and bottom by x^2: \displaystyle \frac{\sin x / x}{(1 - \cos x)/x^2} \to \frac{1}{1/2} = 2.
1. \sin x = \sin(\pi - x). With u = \pi - x, u \to 0 and u \ne 0 when x \ne \pi, so the quotient is \displaystyle \frac{\sin u}{u} \to 1.
0. For x \ne 0, \displaystyle \frac{\sin(x^2)}{x} = x \cdot \frac{\sin(x^2)}{x^2}. With u = x^2, u \to 0 and u \ne 0, so the second factor tends to 1 and the product to 0 \cdot 1 = 0.
\displaystyle \frac{1}{2}. \displaystyle \tan x - \sin x = \frac{\sin x\,(1 - \cos x)}{\cos x}, so the quotient is \displaystyle \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} \cdot \frac{1}{\cos x} \to 1 \cdot \frac{1}{2} \cdot 1.
4. \cos x - \cos 3x = (1 - \cos 3x) - (1 - \cos x). Rescaling Example 5, \displaystyle \frac{1 - \cos 3x}{x^2} = 9 \cdot \frac{1 - \cos 3x}{(3x)^2} \to \frac{9}{2}, and \displaystyle \frac{1 - \cos x}{x^2} \to \frac{1}{2}, so the limit is \displaystyle \frac{9}{2} - \frac{1}{2} = 4.
\displaystyle \frac{\pi}{180}. \tan(x^\circ) = \tan u with \displaystyle u = \frac{\pi x}{180}, so \displaystyle \frac{\tan(x^\circ)}{x} = \frac{\pi}{180} \cdot \frac{\tan u}{u} \to \frac{\pi}{180} \approx 0.01745.
The quotient law needs the bottom limit to be nonzero, and here it is 0, so the law does not apply. The form \displaystyle \frac{0}{0} is not a value and says nothing about whether the limit exists. The squeeze argument shows the limit exists and is 1.
Nothing beyond what the walls allow: they tend to 1 and 2, which differ, so the limit is not determined and need not exist. f(x) = 1 - |x| has limit 1, f(x) = 2 has limit 2, and \displaystyle f(x) = \frac{3}{2} + \frac{1}{2}\cos \frac{1}{x} (with any value from 1 to 2 at 0) stays between 1 and 2 and has no limit at 0.
1. The squeeze theorem needs the trap only at the x \ne 0 in some open interval around 0, and the interval from -0.01 to 0.01 is one. Both walls tend to 1, so \lim_{x \to 0} f(x) = 1; the value f(0) and the behavior farther out play no part.
For x > 0, \displaystyle -\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}, and both walls tend to 0 as x \to \infty, so the limit is 0. The limit 1 is about x \to 0, where \sin x and x are both small and nearly equal; for large x the sine stays between -1 and 1 while x grows without bound.
If k \ne 0, \displaystyle \frac{\sin kx}{x} = k \cdot \frac{\sin kx}{kx} \to k \cdot 1 = k. The substitution is allowed because as x \to 0 with x \ne 0, u = kx \to 0 with u \ne 0, so the values are values of \displaystyle \frac{\sin u}{u} at inputs near 0. If k = 0, \displaystyle \frac{\sin 0}{x} = 0 for every x \ne 0, so the limit is 0 = k; only the rewriting needed k \ne 0.
A sector of x degrees is \displaystyle \frac{x}{360} of the unit circle’s area \pi, so its area is \displaystyle \frac{\pi x}{360}. The chain of areas becomes \displaystyle \frac{1}{2}\sin(x^\circ) < \frac{\pi x}{360} < \frac{1}{2}\tan(x^\circ), and the same steps give \displaystyle \cos(x^\circ) < \frac{\sin(x^\circ)}{\pi x / 180} < 1, so \displaystyle \frac{\sin(x^\circ)}{x} \to \frac{\pi}{180}. Only in radians is the sector’s area half the angle, which is what makes the limit exactly 1.