The Squeeze Theorem

Why this matters¶

How does $\sin x$ compare with $x$ when $x$ is small? A calculator set to radians gives $\sin 0.1 \approx 0.0998334$ , $\sin 0.01 \approx 0.00999983$ and $\sin 0.001 \approx 0.000999999833$ : each is almost exactly $x$ . Physicists use this all the time. For a pendulum that swings through small angles they replace $\sin\theta$ by $\theta$ , which turns an equation they cannot solve into one they can. Why does that work, and how good is the replacement?

The precise question is about the ratio $\frac{\sin x}{x}$ . Its values at the same points are

$x$	0.1	0.01	0.001
$\frac{\sin x}{x}$	0.998334	0.999983	0.9999998

(to the digits shown, and the same at -0.1, -0.01, -0.001). Does the ratio approach 1 as $x \to 0$ ? The limit laws cannot tell us. At $x = 0$ the ratio is “ $\frac{0}{0}$ ”, where the quotient law says nothing (When the laws do not apply), and $\sin x$ is not a polynomial, so there is nothing to factor or cancel.

What we do have is a comparison. The lemma on sine, angle and tangent says that $\sin\theta < \theta < \tan\theta$ for $0 < \theta < \frac{\pi}{2}$ , and we will see that this traps $\frac{\sin x}{x}$ between $\cos x$ and 1. If both of these approach 1, the trapped function has nowhere else to go. This page makes that idea a theorem, the squeeze theorem, and uses it to prove that $\frac{\sin x}{x}$ approaches 1.

Trapping a function¶

Suppose that, near a point $a$ , a function $f$ lies between two functions $g$ and $h$ , and that $g(x)$ and $h(x)$ both approach the same number $L$ as $x \to a$ . Then $f(x)$ , which is caught between them, must approach $L$ too: the two bounds close in on $L$ and leave $f$ no room. This works even when $f$ itself is hard to handle, as long as the bounds are easy.

Here is a function that is hard to handle. As $x$ approaches 0, the factor $\sin(1/x)$ swings between ${-1}$ and 1 faster and faster, and $x \sin(1/x)$ swings with it. But the factor $x$ shrinks the swings: by part (a) of the proposition on the bounds of sine and cosine, $-1 \le \sin(1/x) \le 1$ , so the values of $x\sin(1/x)$ stay between $-\abs{x}$ and $\abs{x}$ .

Figure 1:Graph, for $-0.5 \le x \le 0.5$ and $x \ne 0$ , of one of three functions, chosen by a slider for $k$ , which takes the values ${-1}$ , 0 and 1 and starts at 0. At $k = -1$ it is the lower bound $g(x) = -\abs{x}$ , at $k = 0$ the function $f(x) = x \sin(1/x)$ , and at $k = 1$ the upper bound $h(x) = \abs{x}$ (the widget plots $k\abs{x} + (1 - \abs{k})\,x\sin(1/x)$ ). Taken together, the graphs at $k = -1$ and $k = 1$ are the edges of a band shaped like a bow tie, which narrows to the origin. At $k = 0$ , the graph of $f$ swings between the heights $-\abs{x}$ and $\abs{x}$ , faster and faster as $x$ approaches 0; close to 0 the drawing is a jumble of lines, because the plotted curve joins finitely many computed points. An open circle marks the origin, where the formula is undefined. A table lists the values at $x = 0.1, 0.01, 0.001, -0.001, -0.01, -0.1$ . For $k = 0$ they are -0.0544021, -0.00506366, 0.00082688, 0.00082688, -0.00506366, -0.0544021; for $k = -1$ they are $-\abs{x}$ , and for $k = 1$ they are $\abs{x}$ . At each of these six points the value of $f$ lies between those of $g$ and $h$ . A second slider moves a point along the graph, starting at $x = 0.25$ , with a readout of $x$ and the value there.

Try this: with $k = 0$ , look at the table: do the values of $f$ settle as $x$ gets closer to 0? Now set $k = -1$ and $k = 1$ and compare the three tables row by row. How wide is the band at $x = 0.001$ , and what does that say about $f(0.001)$ ? Then move the traced point towards 0 and switch $k$ between ${-1}$ , 0 and 1 at each position.

The table of $f$ alone looks irregular: its values change sign, and their sizes do not shrink at a steady rate. What the tables show, at the six points they list, is that the value of $f$ lies between those of $g$ and $h$ , in a band whose width $2\abs{x}$ shrinks to 0. A picture and a table only check finitely many points; the squeeze theorem turns the band into a proof that $x\sin(1/x)$ approaches 0 (Example 1).

Main results¶

The squeeze theorem¶

Proof (Rigorous track)

We play the ε–δ game of the definition of a limit for $g$ and $h$ with the same tolerance $\eps$ : near $a$ , both $g(x)$ and $h(x)$ lie strictly between $L - \eps$ and $L + \eps$ , and $f(x)$ , caught between them, lies there too.

Let $\eps > 0$ . Since $\lim_{x \to a} g(x) = L$ , there is a $\delta_1 > 0$ such that $g$ is defined and $\abs{g(x) - L} < \eps$ at every $x$ with $0 < \abs{x - a} < \delta_1$ . Since $\lim_{x \to a} h(x) = L$ , there is a $\delta_2 > 0$ such that $h$ is defined and $\abs{h(x) - L} < \eps$ at every $x$ with $0 < \abs{x - a} < \delta_2$ .

Let $\delta = \min(\delta_1, \delta_2, r)$ , which is positive, and let $0 < \abs{x - a} < \delta$ . Since $\delta$ is at most each of $\delta_1$ , $\delta_2$ and $r$ , also $\abs{x - a} < \delta_1$ , $\abs{x - a} < \delta_2$ and $\abs{x - a} < r$ (property 2 of the order rules, transitivity). So $x$ lies in both windows and in the punctured interval of the hypothesis: $f(x)$ , $g(x)$ and $h(x)$ are defined, $g(x) \le f(x) \le h(x)$ , $\abs{g(x) - L} < \eps$ and $\abs{h(x) - L} < \eps$ . By part (a) of the proposition on distance inequalities, applied to the number $g(x)$ with centre $L$ and radius $\eps$ , the inequality $\abs{g(x) - L} < \eps$ says $L - \eps < g(x) < L + \eps$ ; in the same way, $L - \eps < h(x) < L + \eps$ . Hence

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

(2)

By transitivity (property 2 of the order rules: $u < v \le w$ implies $u < w$ , and $u \le v < w$ implies $u < w$ ), $L - \eps < f(x)$ and $f(x) < L + \eps$ . By part (a) of the proposition on distance inequalities again, this says $\abs{f(x) - L} < \eps$ .

So $f$ is defined at every $x$ with $0 < \abs{x - a} < r$ , which by part (e) of the proposition on distance inequalities is the open interval $(a - r, a + r)$ with $a$ removed; the definition of a limit applies to $f$ , and $\delta$ wins the round $\eps$ . Since $\eps > 0$ was arbitrary, $\lim_{x \to a} f(x) = L$ .

In words. If, near $a$ (but not necessarily at $a$ ), $f$ lies between a lower bound $g$ and an upper bound $h$ that approach the same number $L$ , then $f$ approaches $L$ too. The bounds need to hold only on some punctured interval around $a$ , however small, and they say nothing about $f(a)$ , which may be undefined. The theorem is also called the sandwich theorem or the pinching theorem.

Example. Let $f(x) = x\abs{x}$ . By the definition of the absolute value, $f(x) = x^2$ for $x \ge 0$ and $f(x) = -x^2$ for $x < 0$ . Since $x^2 \ge 0$ , also $-x^2 \le 0 \le x^2$ , so in both cases

-x^2 \le f(x) \le x^2

(3)

for every real $x$ . The bounds are polynomials, so by direct substitution both approach 0 as $x \to 0$ , and the theorem (with any $r$ , say $r = 1$ ) gives $\lim_{x \to 0} x\abs{x} = 0$ . The function $f$ is given by two formulas, and the bounds are single polynomials: the theorem is useful exactly when $f$ is the hard part and the bounds are easy.

Non-example. For $x \ne 0$ , $-1 \le \sin(1/x) \le 1$ , but the bounds ${-1}$ and 1 approach different numbers, so the theorem says nothing about $\lim_{x \to 0} \sin(1/x)$ . In fact that limit does not exist (the example on $\sin(1/x)$ in The Limit of a Function): the bounds leave room for the values to swing between them.

The hypotheses are all needed: Remark 2 gives a counterexample for each. First, a form of the theorem that is often quicker to use. Most squeezes bound the distance from $f(x)$ to $L$ .

Remark 1 (Squeezing the distance)

Let $a$ and $L$ be real numbers, let $r > 0$ , and let $f$ and $b$ be functions that are defined at every $x$ with $0 < \abs{x - a} < r$ , such that $\abs{f(x) - L} \le b(x)$ for every such $x$ . If $\lim_{x \to a} b(x) = 0$ , then $\lim_{x \to a} f(x) = L$ .

Reason. Let $0 < \abs{x - a} < r$ . By part (b) of the proposition on distance inequalities, applied to the number $f(x)$ with centre $L$ and radius $b(x)$ , the inequality $\abs{f(x) - L} \le b(x)$ says

L - b(x) \le f(x) \le L + b(x) .

(4)

By part (a) of the limit laws, $\lim_{x \to a} L = L$ , and $b$ is defined on the open interval $(a - r, a + r)$ except possibly at $a$ , so the difference and sum laws, part (b), give $\lim_{x \to a} \bigl(L - b(x)\bigr) = L - 0 = L$ and $\lim_{x \to a} \bigl(L + b(x)\bigr) = L + 0 = L$ . Both are defined wherever $b$ is. The squeeze theorem, with $g(x) = L - b(x)$ and $h(x) = L + b(x)$ , gives $\lim_{x \to a} f(x) = L$ .

With $g = h$ , the squeeze theorem also says that two functions that are equal on a punctured interval around $a$ have the same limit at $a$ : if $f(x) = g(x)$ for every $x$ with $0 < \abs{x - a} < r$ , then $g(x) \le f(x) \le g(x)$ there, and $\lim_{x \to a} g(x) = L$ gives $\lim_{x \to a} f(x) = L$ . We use this when two formulas agree near $a$ but not everywhere.

Remark 2 (Each hypothesis is needed)

The three counterexamples use one function: $s(x) = \dfrac{\abs{x}}{x}$ for $x \ne 0$ . By the definition of the absolute value, $\abs{x} = x$ for $x > 0$ and $\abs{x} = -x$ for $x < 0$ , so $s(x) = 1$ for $x > 0$ and $s(x) = -1$ for $x < 0$ .

$s$ has no limit at 0. Let $L$ be any real number, and suppose that some $\delta > 0$ wins the round $\eps = 1$ of the definition of a limit. The points $x = \frac{\delta}{2}$ and $x = -\frac{\delta}{2}$ satisfy $0 < \abs{x} = \frac{\delta}{2} < \delta$ , so they lie in the window, and $\abs{1 - L} < 1$ and $\abs{-1 - L} < 1$ . Also $\abs{L - (-1)} = \abs{-1 - L}$ , by property 2 of the absolute value. By part (b) of the triangle inequality, with $x = 1$ , $y = -1$ and $z = L$ ,

2 = \abs{1 - (-1)} \le \abs{1 - L} + \abs{L - (-1)} < 1 + 1 = 2,

(5)

which is impossible. So no $\delta$ wins the round $\eps = 1$ , and $L$ is not the limit. Since $L$ was arbitrary, $s$ has no limit at 0.

The bounds must approach the same number. Let $g(x) = -1$ and $h(x) = 1$ . Then $g(x) \le s(x) \le h(x)$ for every $x \ne 0$ , $\lim_{x \to 0} g(x) = -1$ and $\lim_{x \to 0} h(x) = 1$ , by part (a) of the limit laws, and $s$ has no limit at 0. Bounds with different limits leave room between them.
The inequalities must hold on both sides of $a$ . Let $g(x) = h(x) = 1$ . Then $g(x) \le s(x) \le h(x)$ holds for every $x > 0$ , but for no $x < 0$ , where $s(x) = -1 < 1$ . Both bounds have the limit 1, and $s$ has no limit at 0. The theorem needs the inequalities at every point of a punctured interval $0 < \abs{x - a} < r$ , which reaches both sides of $a$ .
One bound is not enough. Let $g(x) = -1$ . Then $g(x) \le s(x)$ for every $x \ne 0$ and $\lim_{x \to 0} g(x) = -1$ , but $s$ has no limit at 0: a lower bound alone does not stop $f$ from moving up and down above it. In the same way, the upper bound $h(x) = 1$ alone is not enough.

What the theorem does not need: the inequalities at $a$ itself, a value $f(a)$ , or strict inequalities. And its conclusion is no stronger than its hypotheses: if $g(x) < f(x) < h(x)$ , the limit is still only $L$ , with no strict inequality left.

Sine and cosine near 0¶

To use the squeeze theorem for $\frac{\sin x}{x}$ , we need bounds for $\sin x$ and $\cos x$ near 0, and their limits there. Everything rests on the lemma on sine, angle and tangent of Trigonometric Functions, which compares areas and uses no limits, and on the properties of $\sin$ and $\cos$ proved on that page. The lemma is about $0 < \theta < \frac{\pi}{2}$ only. For negative numbers we use the symmetries of property 4 of the point $P(t)$ : $\cos(-t) = \cos t$ and $\sin(-t) = -\sin t$ .

Proof

We reduce a negative $x$ to the positive number $t = \abs{x}$ with the symmetries of sine and cosine, apply the lemma on sine, angle and tangent to $t$ , and get (b) from (a) with the double-angle formula for the cosine.

Step 1: from $x$ to $t = \abs{x}$ . Let $y$ be a real number with $0 < \abs{y} < \frac{\pi}{2}$ , and let $t = \abs{y}$ , so that $0 < t < \frac{\pi}{2}$ . By the definition of the absolute value, $y = t$ if $y > 0$ , and $y = -t$ if $y < 0$ . In the second case, by property 4 of the point $P(t)$ , $\sin y = \sin(-t) = -\sin t$ and $\cos y = \cos(-t) = \cos t$ , and so $\frac{\sin y}{y} = \frac{-\sin t}{-t} = \frac{\sin t}{t}$ . So in both cases

\cos y = \cos t \quad\text{and}\quad \frac{\sin y}{y} = \frac{\sin t}{t},

(6)

and $\sin y$ is $\sin t$ or $-\sin t$ . By property 8 of the point $P(t)$ , $\sin t > 0$ and $\cos t > 0$ , since $0 < t < \frac{\pi}{2}$ . So $\abs{\sin y} = \abs{\sin t} = \sin t > 0$ , by property 2 of the absolute value and the definition of the absolute value.

Now let $0 < \abs{x} < \frac{\pi}{2}$ , and apply step 1 with $y = x$ and $t = \abs{x}$ . By the lemma on sine, angle and tangent,

\sin t < t < \tan t .

(7)

(a) $\abs{\sin x} = \sin t < t = \abs{x}$ .

(c) The number $\frac{1}{t}$ is positive, because $t > 0$ (property 6 of the order rules). Multiplying $\sin t < t$ by it keeps the strict inequality (property 5(a) of the order rules) and gives $\frac{\sin t}{t} < 1$ . Next, $\cos t \ne 0$ , so $\tan t = \frac{\sin t}{\cos t}$ , by the definition of the tangent. The number $\frac{\cos t}{t}$ is positive: multiplying $0 < \cos t$ by the positive number $\frac{1}{t}$ keeps the inequality (property 5(a)). Multiplying $t < \frac{\sin t}{\cos t}$ by $\frac{\cos t}{t}$ keeps the strict inequality (property 5(a)) and gives $\cos t < \frac{\sin t}{t}$ . By step 1,

\cos x = \cos t < \frac{\sin t}{t} = \frac{\sin x}{x} < 1 .

(8)

(b) Let $s = \frac{x}{2}$ , so that $x = 2s$ . By property 4 of the absolute value, $\abs{s} = \frac{\abs{x}}{2}$ , and multiplying $0 < \abs{x} < \frac{\pi}{2}$ by the positive number $\frac12$ (property 5(a) of the order rules) gives $0 < \abs{s} < \frac{\pi}{4}$ ; and $\frac{\pi}{4} < \frac{\pi}{2}$ . So $s$ satisfies the hypothesis of the lemma, and (a), applied to $s$ , gives $\abs{\sin s} < \abs{s}$ . Both sides are non-negative (property 1 of the absolute value), so squaring keeps the strict inequality (property 6 of the absolute value), and by property 5 of the absolute value, $\abs{u}^2 = u^2$ for every real $u$ :

\sin^2 s = \abs{\sin s}^2 < \abs{s}^2 = s^2 = \frac{x^2}{4} .

(9)

Multiplying by the positive number 2 keeps it (property 5(a) of the order rules): $2\sin^2 s < \frac{x^2}{2}$ . By part (f) of the addition formulas, applied to the number $s$ ,

\cos x = \cos 2s = 1 - 2\sin^2 s .

(10)

Adding $1 - 2\sin^2 s - \frac{x^2}{2}$ to both sides of $2\sin^2 s < \frac{x^2}{2}$ keeps it (property 3 of the order rules) and gives $1 - \frac{x^2}{2} < 1 - 2\sin^2 s = \cos x$ . For the other inequality, step 1 with $y = s$ gives $\abs{\sin s} > 0$ ; multiplying $0 < \abs{\sin s}$ by the positive number $\abs{\sin s}$ (property 5(a)) gives $0 < \abs{\sin s}^2 = \sin^2 s$ , and multiplying by 2 gives $0 < 2\sin^2 s$ . Adding $1 - 2\sin^2 s$ to both sides (property 3) gives $\cos x = 1 - 2\sin^2 s < 1$ .

The inequalities of the lemma are what the squeeze theorem needs. With them we get the limits of $\sin$ and $\cos$ at 0, without knowing that $\sin$ and $\cos$ are continuous.

Proof

We prove (a) from the definition, and then squeeze $\sin x$ and $\cos x$ with the bounds of Lemma 1.

(a) The function $\abs{x}$ is defined for every real $x$ . Let $\eps > 0$ and take $\delta = \eps$ . For every real $x$ , $\abs{x} \ge 0$ (property 1 of the absolute value), so $\bigl\lvert \abs{x} - 0 \bigr\rvert = \bigl\lvert \abs{x} \bigr\rvert = \abs{x}$ , by the definition of the absolute value. So if $0 < \abs{x - 0} < \delta$ , then $\bigl\lvert \abs{x} - 0 \bigr\rvert = \abs{x} < \eps$ . Hence $\lim_{x \to 0} \abs{x} = 0$ .

(b) The functions $\sin x$ and $\abs{x}$ are defined for every real $x$ . For every $x$ with $0 < \abs{x} < \frac{\pi}{2}$ , part (a) of the lemma on sine and cosine near 0 gives $\abs{\sin x - 0} = \abs{\sin x} < \abs{x}$ , so in particular $\abs{\sin x - 0} \le \abs{x}$ . By part (a) above, $\lim_{x \to 0} \abs{x} = 0$ . So Remark 1, with $a = 0$ , $L = 0$ , $r = \frac{\pi}{2}$ , $f(x) = \sin x$ and $b(x) = \abs{x}$ , gives $\lim_{x \to 0} \sin x = 0$ .

(c) For every $x$ with $0 < \abs{x} < \frac{\pi}{2}$ , part (b) of the lemma on sine and cosine near 0 gives

1 - \frac{x^2}{2} \le \cos x \le 1 .

(11)

The lower bound is a polynomial, so by direct substitution, part (a), $\lim_{x \to 0} \bigl(1 - \frac{x^2}{2}\bigr) = 1 - \frac{0^2}{2} = 1$ . The upper bound is the constant 1, and $\lim_{x \to 0} 1 = 1$ by part (a) of the limit laws. All three functions are defined for every real $x$ . The squeeze theorem, with $a = 0$ , $L = 1$ , $r = \frac{\pi}{2}$ , $g(x) = 1 - \frac{x^2}{2}$ , $f(x) = \cos x$ and $h(x) = 1$ , gives $\lim_{x \to 0} \cos x = 1$ .

So the limits of $\sin$ and $\cos$ at 0 are their values there, $\sin 0 = 0$ and $\cos 0 = 1$ . That is no surprise, but it needed a proof: neither function is a polynomial, so direct substitution does not apply to them.

The limit of sine x over x¶

Proof

We squeeze $\frac{\sin x}{x}$ between $\cos x$ and 1, which both approach 1; the bounds come from comparing areas, through the lemma on sine, angle and tangent, and use no derivatives and no continuity of $\sin$ or $\cos$ .

The function $\frac{\sin x}{x}$ is defined at every $x \ne 0$ , and $\cos x$ and the constant 1 at every real $x$ . For every $x$ with $0 < \abs{x} < \frac{\pi}{2}$ , part (c) of the lemma on sine and cosine near 0 gives $\cos x < \frac{\sin x}{x} < 1$ , and so

\cos x \le \frac{\sin x}{x} \le 1 .

(13)

By part (c) of the lemma on limits at 0, $\lim_{x \to 0} \cos x = 1$ , and $\lim_{x \to 0} 1 = 1$ by part (a) of the limit laws. The squeeze theorem, with $a = 0$ , $L = 1$ , $r = \frac{\pi}{2}$ , $g(x) = \cos x$ , $f(x) = \frac{\sin x}{x}$ and $h(x) = 1$ , gives $\lim_{x \to 0} \frac{\sin x}{x} = 1$ .

The chain of reasoning, from the start: areas of a triangle, a sector and a larger triangle give $\sin\theta < \theta < \tan\theta$ (Lemma 1); dividing by $\theta$ and using the symmetries of $\sin$ and $\cos$ traps $\frac{\sin x}{x}$ between $\cos x$ and 1 (Lemma 1); the double-angle formula squeezes $\cos x$ to 1 (Lemma 2); and the squeeze theorem finishes. No step uses a derivative, a series, or the continuity of $\sin$ or $\cos$ .

Why radians. The lemma compares $\sin\theta$ with $\theta$ , the length of an arc of the unit circle, and that length is the measure of the angle in radians (Definition 1). An angle of $x$ degrees is $\frac{\pi x}{180}$ radians, so in degrees the ratio is $\frac{\sin(cx)}{x}$ with $c = \frac{\pi}{180}$ . At $x = 0.01$ it is about 0.0174533, close to $\frac{\pi}{180} \approx 0.0174533$ rather than to 1. The lemmas prove that this is the limit, with no new tool.

Let $c > 0$ , and let $0 < \abs{x} < \frac{\pi}{2c}$ . By property 4 of the absolute value, $\abs{cx} = \abs{c}\,\abs{x} = c\abs{x}$ (since $c > 0$ , $\abs{c} = c$ by the definition of the absolute value), and multiplying $0 < \abs{x} < \frac{\pi}{2c}$ by the positive number $c$ keeps both inequalities (property 5(a) of the order rules): $0 < \abs{cx} < \frac{\pi}{2}$ . So parts (b) and (c) of the lemma on sine and cosine near 0 apply to the number $cx$ :

1 - \frac{c^2x^2}{2} < \cos(cx) < \frac{\sin(cx)}{cx} < 1 .

(14)

By transitivity (property 2 of the order rules), $1 - \frac{c^2x^2}{2} < \frac{\sin(cx)}{cx} < 1$ . Multiplying by the positive number $c$ keeps both inequalities (property 5(a)), and $c \cdot \frac{\sin(cx)}{cx} = \frac{\sin(cx)}{x}$ , so

c - \frac{c^3x^2}{2} < \frac{\sin(cx)}{x} < c ,

(15)

and so the same with $\le$ . The lower bound is a polynomial in $x$ , so by part (a) of direct substitution it approaches $c - 0 = c$ ; the upper bound is the constant $c$ , which approaches $c$ by part (a) of the limit laws. The bounds are defined for every real $x$ , and $\frac{\sin(cx)}{x}$ for every $x \ne 0$ . The number $r = \frac{\pi}{2c}$ is positive: it is the positive number $\frac{\pi}{2}$ times $\frac{1}{c} > 0$ (properties 6 and 5(a) of the order rules). So the squeeze theorem, with $a = 0$ , $L = c$ and this $r$ , gives

\lim_{x \to 0} \frac{\sin(cx)}{x} = c .

(16)

With $c = \frac{\pi}{180}$ : in degrees, the ratio approaches $\frac{\pi}{180}$ , not 1.

Proof

We multiply numerator and denominator by $1 + \cos x$ , which turns $1 - \cos x$ into $\sin^2 x$ , write the result as a product of $\frac{\sin x}{x}$ and a quotient, and apply the limit laws.

Rewriting. Let $0 < \abs{x} < \frac{\pi}{2}$ . Then $\cos x = \cos\abs{x}$ : for $x > 0$ because $\abs{x} = x$ , and for $x < 0$ because $\abs{x} = -x$ and $\cos(-x) = \cos x$ , by property 4 of the point $P(t)$ . Since $0 < \abs{x} < \frac{\pi}{2}$ , property 8 gives $\cos\abs{x} > 0$ . So $\cos x > 0$ , $1 + \cos x > 1 > 0$ (property 3 of the order rules) and $1 + \cos x \ne 0$ . Multiplying numerator and denominator by $1 + \cos x$ , and using the Pythagorean identity, $1 - \cos^2 x = \sin^2 x$ ,

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

(18)

Call the last product $F(x)$ . It is defined at every $x$ with $0 < \abs{x} < \frac{\pi}{2}$ , and equal to $\frac{1 - \cos x}{x}$ there.

The limits of the parts. By the lemma on limits at 0, $\lim_{x \to 0} \sin x = 0$ and $\lim_{x \to 0} \cos x = 1$ ; with $\lim_{x \to 0} 1 = 1$ (part (a) of the limit laws), the sum law, part (b), gives $\lim_{x \to 0} (1 + \cos x) = 2$ . These functions are defined for every real $x$ . Since $2 \ne 0$ , the quotient law, part (d) of the limit laws, applies with $f(x) = \sin x$ , $g(x) = 1 + \cos x$ , $L = 0$ and $M = 2$ , and gives

\lim_{x \to 0} \frac{\sin x}{1 + \cos x} = \frac{0}{2} = 0 ,

(19)

and it says that this quotient is defined on an open interval around 0. By Theorem 2, $\lim_{x \to 0} \frac{\sin x}{x} = 1$ , and $\frac{\sin x}{x}$ is defined at every $x \ne 0$ . So the product law, part (c) of the limit laws, gives $\lim_{x \to 0} F(x) = 1 \cdot 0 = 0$ .

Back to the original function. $\frac{1 - \cos x}{x}$ is defined at every $x \ne 0$ , and for $0 < \abs{x} < \frac{\pi}{2}$ it equals $F(x)$ , so $F(x) \le \frac{1 - \cos x}{x} \le F(x)$ there. The squeeze theorem, with $g = h = F$ , $L = 0$ and $r = \frac{\pi}{2}$ , gives $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ .

The only law used with a quotient was the quotient law with $M = 2$ . The original quotient $\frac{1 - \cos x}{x}$ has $M = 0$ in its denominator, so the quotient law cannot be applied to it directly; the rewriting moves the “ $\frac{0}{0}$ ” into $\frac{\sin x}{x}$ , whose limit we know.

Worked examples¶

Example 1 (An oscillating factor:

\lim_{x \to 0} x \sin(1/x)

)

Find $\displaystyle \lim_{x \to 0} x\sin\Bigl(\frac{1}{x}\Bigr)$ , the function of the squeeze-band figure.

Why the laws fail. The product law would need a limit of $\sin(1/x)$ at 0, and the bound $-1 \le \sin(1/x) \le 1$ gives none. We bound the distance to the guess 0 instead.
The bound. Let $x \ne 0$ . By part (a) of the proposition on the bounds of sine and cosine, with $t = \frac{1}{x}$ , $-1 \le \sin(1/x) \le 1$ , which by part (b) of the proposition on distance inequalities (centre 0, radius 1) says $\abs{\sin(1/x)} \le 1$ . By property 4 of the absolute value, and multiplying $\abs{\sin(1/x)} \le 1$ by the number $\abs{x} \ge 0$ , which gives only $\le$ (property 5(b) of the order rules),
$\Bigl\lvert x\sin\Bigl(\frac{1}{x}\Bigr) - 0\Bigr\rvert = \abs{x}\,\Bigl\lvert\sin\Bigl(\frac{1}{x}\Bigr)\Bigr\rvert \le \abs{x} .$
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Squeeze. Both $x\sin(1/x)$ and $\abs{x}$ are defined at every $x \ne 0$ , and $\lim_{x \to 0} \abs{x} = 0$ by part (a) of the lemma on limits at 0. By Remark 1, with $a = 0$ , $L = 0$ , $r = 1$ and $b(x) = \abs{x}$ , the limit is 0.

\boxed{\lim_{x \to 0} x\sin\Bigl(\frac{1}{x}\Bigr) = 0}

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Check. The table of the figure, at $k = 0$ : $\abs{-0.0544021} \le 0.1$ , $\abs{-0.00506366} \le 0.01$ and $0.00082688 \le 0.001$ . ✓ At $x = \frac{1}{\pi}$ , where $\sin(1/x) = \sin\pi = 0$ , the value is 0; at $x = \frac{2}{\pi}$ , where $\sin\frac{\pi}{2} = 1$ , the value is $\frac{2}{\pi}$ , on the upper edge of the band: the bound cannot be improved in general. ✓

Example 2 (A related trigonometric limit:

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

)

Find $\displaystyle \lim_{x \to 0} \frac{\sin 3x}{x}$ .

Strategy. Substituting $x = 0$ gives “ $\frac{0}{0}$ ”. The argument of the sine is $3x$ , not $x$ , so Theorem 2 does not apply as it stands. We write $\sin 3x$ in terms of $\sin x$ and $\cos x$ with the addition formulas, so that $\frac{\sin x}{x}$ appears.
Rewrite $\sin 3x$ . For every real $x$ , by parts (c), (e) and (f) of the addition formulas, with $s = 2x$ and $t = x$ in (c), and then the Pythagorean identity, $\cos^2 x = 1 - \sin^2 x$ :
$\begin{aligned} \sin 3x &= \sin 2x \cos x + \cos 2x \sin x \\ &= 2\sin x \cos^2 x + (1 - 2\sin^2 x)\sin x \\ &= 2\sin x\,(1 - \sin^2 x) + \sin x - 2\sin^3 x \\ &= 3\sin x - 4\sin^3 x . \end{aligned}$
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Divide by $x$ . For every $x \ne 0$ ,
$\frac{\sin 3x}{x} = \frac{\sin x}{x}\,\bigl(3 - 4\sin^2 x\bigr) .$
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Both sides are defined at exactly the $x \ne 0$ , so they are the same function.
The limits of the factors. By Theorem 2, the first factor approaches 1. By part (b) of the lemma on limits at 0, $\lim_{x \to 0} \sin x = 0$ , so the power law, part (e) of the limit laws with $n = 2$ , gives $\lim_{x \to 0} \sin^2 x = 0$ ; the “in particular” of part (c), with $c = 4$ , gives $\lim_{x \to 0} 4\sin^2 x = 0$ ; and part (a) and the difference law, part (b), give $\lim_{x \to 0} (3 - 4\sin^2 x) = 3 - 0 = 3$ . All these functions are defined for every real $x$ .
Combine. By the product law, part (c), $\lim_{x \to 0} \frac{\sin 3x}{x} = 1 \cdot 3 = 3$ .

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

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Check. At $x = 0.01$ : $\frac{\sin 0.03}{0.01} \approx 2.99955$ , close to 3. ✓ The identity of step 2 at $x = \frac{\pi}{2}$ : $\sin\frac{3\pi}{2} = -1$ , and $3\sin\frac{\pi}{2} - 4\sin^3\frac{\pi}{2} = 3 - 4 = -1$ . ✓

Example 3 (How small is a small angle?)

A model of a pendulum replaces $\sin\theta$ by $\theta$ , where $\theta$ is the angle of the pendulum from the vertical, in radians. The relative error of this replacement is $\frac{\theta - \sin\theta}{\theta} = 1 - \frac{\sin\theta}{\theta}$ , for $\theta \ne 0$ . For which angles do the bounds on this page guarantee a relative error below $1\,\%$ , that is, below 0.01?

A bound for the error. Let $0 < \abs{\theta} < \frac{\pi}{2}$ . By parts (b) and (c) of the lemma on sine and cosine near 0,
$1 - \frac{\theta^2}{2} < \cos\theta < \frac{\sin\theta}{\theta} < 1 .$
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By transitivity (property 2 of the order rules), $1 - \frac{\theta^2}{2} < \frac{\sin\theta}{\theta} < 1$ . Adding $\frac{\theta^2}{2} - \frac{\sin\theta}{\theta}$ to the first inequality and $-\frac{\sin\theta}{\theta}$ to the second keeps them (property 3 of the order rules):
$0 < 1 - \frac{\sin\theta}{\theta} < \frac{\theta^2}{2} .$
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So the relative error is positive and less than $\frac{\theta^2}{2}$ : $\theta$ is larger than $\sin\theta$ in absolute value (for $\theta > 0$ , larger), as part (a) of the same lemma, $\abs{\sin\theta} < \abs{\theta}$ , also says.
When is the bound at most 0.01? $\frac{\theta^2}{2} \le 0.01$ exactly when $\theta^2 \le 0.02$ (multiplying by the positive number 2, or by $\frac12$ , keeps inequalities, property 5(b) of the order rules). By property 5 of the absolute value, $\theta^2 = \abs{\theta}^2$ , and $0.02 = \bigl(\sqrt{0.02}\bigr)^2$ (the square-root remark). Both $\abs{\theta}$ and $\sqrt{0.02}$ are non-negative, so by property 6 of the absolute value this holds exactly when $\abs{\theta} \le \sqrt{0.02} \approx 0.141421$ (property 6 gives this for $<$ and for $=$ ; the two parts together give it for $\le$ ).
Conclusion. Step 1 needs $\abs{\theta} < \frac{\pi}{2}$ , so we compare $\sqrt{0.02}$ with $\frac{\pi}{2}$ , without a decimal value of $\pi$ . By the proposition bounding $\pi$ in Trigonometric Functions, $2\sqrt{2} < \pi$ . Multiplying by the positive number $\frac12$ keeps the inequality (property 5(a) of the order rules): $\sqrt{2} < \frac{\pi}{2}$ . Next, $\sqrt{0.02}$ and $\sqrt{2}$ are non-negative, their squares are 0.02 and 2, and $0.02 < 2$ ; so $\sqrt{0.02} < \sqrt{2}$ , because squaring keeps the order of non-negative numbers (the square-root remark). By transitivity (property 2 of the order rules), $\sqrt{0.02} < \frac{\pi}{2}$ . So every $\theta$ with $0 < \abs{\theta} \le \sqrt{0.02}$ has $\abs{\theta} < \frac{\pi}{2}$ (transitivity again), step 1 applies to it, and then $1 - \frac{\sin\theta}{\theta} < \frac{\theta^2}{2} \le 0.01$ (transitivity). In degrees, $\sqrt{0.02}$ radians is $\frac{180\sqrt{0.02}}{\pi} \approx 8.10$ degrees, since a full turn is $2\pi$ radians (Definition 1) and 360 degrees.

\boxed{0 < \abs{\theta} \le \sqrt{0.02} \approx 0.141}

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in radians, about 8.1 degrees: for such angles, replacing $\sin\theta$ by $\theta$ makes a relative error of less than $1\,\%$ .

Check. At $\theta = 0.14$ : $\sin 0.14 \approx 0.139543$ , so $1 - \frac{\sin 0.14}{0.14} \approx 0.00326$ , less than $\frac{0.14^2}{2} = 0.0098$ , which is less than 0.01. ✓ Units: the angle is in radians, the relative error has no unit. ✓ The bound is not sharp: at $\theta = \sqrt{0.02}$ the relative error is about 0.00333, a third of 0.01. The bounds guarantee $1\,\%$ ; they do not say where the error reaches it.

Common mistakes¶

Using the product law with a bounded factor

✗ Wrong: “ $\lim_{x \to 0} x\sin(1/x) = \lim_{x \to 0} x \cdot \lim_{x \to 0} \sin(1/x) = 0$ , because $\lim_{x \to 0} x = 0$ and $\sin(1/x)$ is between ${-1}$ and 1.”

Why: the product law, part (c) of the limit laws, needs both factors to have limits, as real numbers. “Between ${-1}$ and 1” is a bound, not a limit, and $\sin(1/x)$ has no limit at 0 (the example on $\sin(1/x)$ in The Limit of a Function). So $\lim_{x \to 0} \sin(1/x)$ cannot be written down at all.

✓ Right: squeeze. Since $\bigl\lvert x\sin(1/x)\bigr\rvert \le \abs{x}$ for $x \ne 0$ , and $\abs{x} \to 0$ , the limit is 0 (Example 1). The bound is used, but through the squeeze theorem, not the product law.

“

\frac{\sin(\text{anything})}{\text{anything}}

” with two different anythings

✗ Wrong: “ $\displaystyle \lim_{x \to 0} \frac{\sin 3x}{x} = 1$ , by the limit of $\frac{\sin x}{x}$ .”

Why: Theorem 2 is about $\frac{\sin x}{x}$ , with the same $x$ in the sine and in the denominator. In $\frac{\sin 3x}{x}$ they differ, and the values tell: at $x = 0.01$ the quotient is about 2.99955, not close to 1.

✓ Right: rewrite until $\frac{\sin x}{x}$ appears, then use the limit laws: $\frac{\sin 3x}{x} = \frac{\sin x}{x}\,(3 - 4\sin^2 x) \to 1 \cdot 3 = 3$ (Example 2).

Rigorous track¶

The proof of the squeeze theorem is in the dropdown under its statement. Two more remarks for readers who want the details.

Why not just use “

\sin

is continuous”?

A tempting proof of Lemma 2 is: “ $\sin$ and $\cos$ are continuous, so their limits at 0 are their values there.” That is true, but on this site it is circular. Continuity of $\sin$ and $\cos$ is proved from these two limits at 0 (with the addition formulas), so it cannot be used to prove them. The same goes for any argument through the derivative of $\sin$ , or through a power series for $\sin$ : the derivative is computed from Theorem 2, and this course defines $\sin$ by the unit circle, not by a series. The only facts about $\sin$ and $\cos$ used on this page are the statements of Trigonometric Functions: the lemma on sine, angle and tangent, the properties of the point $P(t)$ , the definition of the tangent, the Pythagorean identity, the addition formulas, the bounds $-1 \le \sin t, \cos t \le 1$ , and the bound $2\sqrt{2} < \pi$ .

The squeeze theorem with “eventually” bounds

The theorem asks for the inequalities $g(x) \le f(x) \le h(x)$ only on some punctured interval $0 < \abs{x - a} < r$ ; what happens farther from $a$ does not matter. For example, $\abs{\sin x} < \abs{x}$ was proved only for $0 < \abs{x} < \frac{\pi}{2}$ , and that was enough for $\lim_{x \to 0} \sin x = 0$ . In quantifiers, the hypothesis reads

\begin{aligned} &\exists r > 0 \ \ \forall x \colon \\ &\quad 0 < \abs{x - a} < r \implies g(x) \le f(x) \le h(x), \end{aligned}

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and the proof chooses $\delta = \min(\delta_1, \delta_2, r)$ so that the window lies inside the interval where the inequalities hold.

Summary¶

Squeeze theorem (Theorem 1): if $g(x) \le f(x) \le h(x)$ on a punctured interval around $a$ , and $g$ and $h$ both approach $L$ , then
$\boxed{\lim_{x \to a} f(x) = L .}$
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Both bounds must approach the same $L$ , and the inequalities must hold on both sides of $a$ (Remark 2).
To show $f(x) \to L$ , it is often enough to bound the distance: $\abs{f(x) - L} \le b(x)$ with $b(x) \to 0$ (Remark 1).
To handle a bounded factor such as $\sin(1/x)$ , which has no limit, bound the distance, as in $\bigl\lvert x\sin(1/x)\bigr\rvert \le \abs{x}$ with $\abs{x} \to 0$ , which gives $x\sin(1/x) \to 0$ (Example 1); the product law does not apply there, the squeeze theorem does.
Near 0, $\abs{\sin x} < \abs{x}$ and $1 - \frac{x^2}{2} < \cos x < 1$ , so $\sin x \to 0$ and $\cos x \to 1$ (Lemma 1, Lemma 2).
In radians, $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1$ (Theorem 2) and $\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ (Corollary 1), proved from areas and the squeeze theorem, without derivatives.
Other trigonometric limits: rewrite until $\frac{\sin x}{x}$ appears, then use the limit laws.

Exercises¶

Where this leads¶

These topics build on this one:

Continuity of Elementary Functions (coming soon)
Derivatives of Trigonometric Functions (coming soon)

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The squeeze theorem is the tool for limits that the laws cannot reach: oscillating factors, and the limits of $\sin$ and $\cos$ , on which the calculus of the trigonometric functions rests.

Why this matters¶

Trapping a function¶

Main results¶

The squeeze theorem¶

Sine and cosine near 0¶

The limit of sine x over x¶

A related limit¶

Worked examples¶

Common mistakes¶

Rigorous track¶

Summary¶

Exercises¶

Where this leads¶