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Limits by algebra

Limits pass through sums, products and quotients, so most limits come from substituting. When substitution gives 0/0, factor and cancel, rationalise, or squeeze: areas on the unit circle trap sin(x)/x and show that it tends to 1.
M04.1.2L1 Core25 minIn review

Before this lesson

M04.1.1 What a limit is M01.5 Trigonometry and the unit circle

After this lesson you can

  • State the limit laws for sums, products and quotients, and prove the sum law from the definition.
  • Find limits of polynomials and rational functions by direct substitution, and say when it is allowed.
  • Resolve a 0/0 form by factoring and cancelling, or by rationalising.
  • Use the squeeze theorem with areas on the unit circle to show that sin(x)/x tends to 1 as x tends to 0.
  • Check a limit found by hand against a table of values in code.

In review. The maths and code on this page are checked by automated tests on every change, but an independent reader has not signed it off yet. Spot a mistake or a confusing step? Tell us.

The question

The last lesson proved two limits by playing the tolerance game. Training a network needs the slope of every function it is built from, and every slope is a limit, so calculus needs a faster way. Most limits fall to a few laws in one line. One does not: the slope of the sine curve at 0 comes down to \({\sin(x)/x}\) as \(x\) tends to 0, which is \({0/0}\) and has nothing to cancel. How are limits found in practice, and how is that one pinned down?

From the last lesson

1. Does the value \({f(a)}\) affect \({\lim_{x \to a} f(x)}\)?

Show answer

No. The definition only looks at \(x\) with \({0 < \abs{x - a}}\), so \({f(a)}\) may differ from the limit or not exist at all.

2. For \({f(x) = 2x + 1}\) at \({x = 1}\), which window \(\delta\) answers a tolerance \(\varepsilon\)?

Show answer

\({\delta = \varepsilon / 2}\), because \({\abs{(2x + 1) - 3} = 2\,\abs{x - 1}}\).

Intuition

If \({f(x)}\) is near \(\ell\) and \({g(x)}\) is near \(m\), then \({f(x) + g(x)}\) is near \({\ell + m}\), the product is near \({\ell m}\), and the quotient is near \({\ell / m}\), as long as \(m\) is not 0. Limits follow the arithmetic of their parts. So for a function built from \(x\) and numbers by adding, multiplying and dividing, the limit at \(a\) is what you get by putting \(a\) in.

Putting \(a\) in fails when the bottom of a fraction becomes 0. If the top does not, the function blows up there, which is the next lesson’s subject. If the top is 0 as well, substitution gives \({0/0}\). That is not an answer. It is a signal that something in the formula is hiding the value, and there are two tools for finding it.

Two graphs. Left, titled Cancel the hidden factor: the blue curve of x squared plus x minus 2, over x squared minus 1, falling from about 2 at x equals 0 to about 1.3 at x equals 2.7, with a hollow circle at (1, 1.5); a black dashed line at height 1.5 is labelled limit three halves. Right, titled Squeeze it between two: for x from minus 1.6 to 1.6, a sky-blue horizontal line at height 1 labelled upper bound 1 and a sky-blue arch, the graph of cosine x, labelled lower bound cos x, with the band between them shaded; the blue graph of sine x over x runs inside the band and has a hollow circle at (0, 1), where both bounds meet.
M04.1.2-F1

Two tools for \({0/0}\). Left: the factor \({x - 1}\) is hidden in the top and the bottom; cancel it and the graph is \({(x + 2)/(x + 1)}\) with one point missing, at height \({\tfrac32}\). Right: nothing cancels in \({\sin(x)/x}\), but it is trapped between \({\cos x}\) and 1, which both head to 1.

Build it up

Step 1. The two simplest limits. A constant stays put, so \({\lim_{x \to a} c = c}\): every window works. And \({\lim_{x \to a} x = a}\): the output is the input, so the window \({\delta = \varepsilon}\) works. Everything else is built from these two.

Step 2. Sums: the gaps add. Suppose \({f(x)}\) is within \({\varepsilon / 2}\) of \(\ell\) and \({g(x)}\) is within \({\varepsilon / 2}\) of \(m\). Each gap is a distance on the number line. Laid end to end, the two gaps reach less than \({\varepsilon / 2 + \varepsilon / 2 = \varepsilon}\), so \({f(x) + g(x)}\) is within \(\varepsilon\) of \({\ell + m}\). If the gaps point opposite ways they partly cancel, which only helps.

Three horizontal number lines, one above the other. Top, labelled f of x near ell: an orange dashed band of half-width epsilon over 2 round a black tick marked ell, and a blue dot inside the band, joined to the tick by a short reddish-purple bar labelled gap. Middle, labelled g of x near m: the same picture round a tick marked m. Bottom, labelled f of x plus g of x near ell plus m: a band twice as wide, of half-width epsilon, round a tick marked ell plus m; the two reddish-purple gaps are laid end to end from the tick, labelled the two gaps, end to end, and the blue dot at their end is still inside the band.
M04.1.2-F2

The sum law in one picture. Each gap is less than \({\varepsilon / 2}\), so the two gaps together are less than \(\varepsilon\), and the sum lands inside a band of half-width \(\varepsilon\) round \({\ell + m}\).

Step 3. Products, quotients and powers. The same kind of argument, with more care about sizes, gives products and quotients; we state those laws without proof. A power is a repeated product, so \({\lim_{x \to a} x^n = a^n}\).

Step 4. Substitution. A polynomial is built from \(x\) and numbers by multiplying and adding, so the laws give its limit by putting \(a\) in. For the course’s cubic \({g(x) = x^3 - 3x}\),

\[ \begin{aligned} \lim_{x \to 2} (x^3 - 3x) &= 2^3 - 3 \times 2 \\ &= 8 - 6 = 2 . \end{aligned} \]

The formal statement

NoteThe limit laws

Suppose \({\lim_{x \to a} f(x) = \ell}\) and \({\lim_{x \to a} g(x) = m}\), and let \(c\) be a number. Then, as \(x\) tends to \(a\),

  • sum and difference: \({f(x) \pm g(x)}\) tends to \({\ell \pm m}\);
  • constant multiple: \({c\,f(x)}\) tends to \({c\,\ell}\);
  • product: \({f(x)\,g(x)}\) tends to \({\ell\,m}\);
  • quotient: \({f(x)/g(x)}\) tends to \({\ell / m}\), provided \({m \neq 0}\);
  • power and root: \({f(x)^n}\) tends to \({\ell^{\,n}}\) for a whole number \({n \geq 1}\), and \({\sqrt{f(x)}}\) tends to \({\sqrt{\ell}}\) when \({f(x) \geq 0}\) near \(a\).

Proof of the sum law. Let \({\varepsilon > 0}\). The definition of a limit holds for every positive tolerance, so use it with \({\varepsilon / 2}\). There is a \({\delta_1 > 0}\) such that \({\abs{f(x) - \ell} < \varepsilon / 2}\) whenever \({0 < \abs{x - a} < \delta_1}\), and a \({\delta_2 > 0}\) such that \({\abs{g(x) - m} < \varepsilon / 2}\) whenever \({0 < \abs{x - a} < \delta_2}\). Let \(\delta\) be the smaller of \(\delta_1\) and \(\delta_2\). If \({0 < \abs{x - a} < \delta}\), both bounds hold, and

\[ \begin{aligned} &\abs{(f(x) + g(x)) - (\ell + m)} \\ &\quad= \abs{(f(x) - \ell) + (g(x) - m)} \\ &\quad\leq \abs{f(x) - \ell} + \abs{g(x) - m} \\ &\quad< \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon . \end{aligned} \]

The second line only regroups. The third is the triangle inequality, \({\abs{u + v} \leq \abs{u} + \abs{v}}\): two moves along the number line never take you further than their lengths added. So a window exists for every \(\varepsilon\), and the sum tends to \({\ell + m}\).

Direct substitution. For a polynomial \(p\), \({\lim_{x \to a} p(x) = p(a)}\). For a rational function \({p(x)/q(x)}\) with \({q(a) \neq 0}\), the limit is \({p(a)/q(a)}\). Both follow by building the function from \(x\) and constants with the laws. When \({q(a) = 0}\) the quotient law is silent, and the work starts.

NoteThe squeeze theorem

Suppose \({g(x) \leq f(x) \leq h(x)}\) for every \(x\) near \(a\) (except perhaps \(a\) itself), and \(g\) and \(h\) both tend to \(\ell\) as \(x\) tends to \(a\). Then \(f\) tends to \(\ell\) too.

Why. Given \({\varepsilon > 0}\), take a window small enough for both \(g\) and \(h\) to be within \(\varepsilon\) of \(\ell\), and inside the range where the inequalities hold. There

\[ \begin{aligned} \ell - \varepsilon &< g(x) \leq f(x) \\ &\leq h(x) < \ell + \varepsilon , \end{aligned} \]

so \(f(x)\) is trapped in the band too.

The limit of \({\sin(x)/x}\). Take \({0 < x < \pi/2}\) and the circle of radius 1. Mark the centre O, the point \({A = (1, 0)}\), the point \({P = (\cos x, \sin x)}\) at angle \(x\), and the point \({T = (1, \tan x)}\) where the line OP meets the upright line through A.

Three copies of a quarter of the unit circle, with centre O, the point A at (1, 0) on the horizontal axis, a point P on the circle at angle x, and the line from O through P extended to T, directly above A. In the first copy the triangle O A P is shaded, labelled triangle OAP: one half sine x. In the second the pie-slice sector O A P, bounded by the arc, is shaded, labelled sector OAP: x over 2. In the third the larger triangle O A T is shaded, labelled triangle OAT: one half tangent x. Less-than-or-equal signs stand between the three copies.
M04.1.2-F3

Three nested shapes. The triangle OAP fits inside the sector OAP, which fits inside the triangle OAT, so their areas come in that order.

Find the three areas.

  • The triangle OAP has base \({OA = 1}\) and height \({\sin x}\), so its area is \({\tfrac12 \sin x}\).
  • The disc of radius 1 has area \(\pi\) (the school formula \({\pi r^2}\) with \({r = 1}\)). Angles are in radians (M01.5.3), so a full turn is \({2\pi}\), and a sector with angle \(x\) is the fraction \({x / (2\pi)}\) of the disc. Equal angles cut equal slices, so the area is shared in proportion to the angle, and the sector’s area is \({\tfrac{x}{2\pi} \times \pi = \tfrac{x}{2}}\).
  • The triangle OAT has base 1 and height \({\tan x}\), so its area is \({\tfrac12 \tan x}\).

The shapes are nested, so the areas are in order. Doubling them,

\[ \sin x \;\leq\; x \;\leq\; \tan x . \]

Divide by \({\sin x}\), which is positive, and use \({\tan x = \sin x / \cos x}\):

\[ 1 \;\leq\; \frac{x}{\sin x} \;\leq\; \frac{1}{\cos x} . \]

All three are positive, and taking reciprocals reverses the order:

\[ \cos x \;\leq\; \frac{\sin x}{x} \;\leq\; 1 . \]

For \({-\pi/2 < x < 0}\) the same holds, because \({\sin(-x)/(-x) = \sin(x)/x}\) and \({\cos(-x) = \cos x}\). Last, \({\cos x}\) tends to 1. From \({\sin x \leq x}\) and the symmetry, \({\abs{\sin x} \leq \abs{x}}\), and since \({\cos x > 0}\) here,

\[ \begin{aligned} 0 \leq 1 - \cos x &= \frac{1 - \cos^2 x}{1 + \cos x} \\ &= \frac{\sin^2 x}{1 + \cos x} \\ &\leq \sin^2 x \;\leq\; x^2 . \end{aligned} \]

So \({1 - \cos x}\) is squeezed between 0 and \({x^2}\), which both tend to 0, and \({\cos x}\) tends to 1. Both bounds on \({\sin(x)/x}\) head to 1, and the squeeze theorem gives

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 . \]

This is the limit that M04.3.2 uses to differentiate the sine.

Worked example

Find three limits: \({x^3 - 3x}\) as \(x\) tends to 2, \({(x^2 + x - 2)/(x^2 - 1)}\) as \(x\) tends to 1, and \({(\sqrt{x} - 3)/(x - 9)}\) as \(x\) tends to 9.

The cubic. A polynomial: substitute, as in Step 4. The limit is \({8 - 6 = 2}\).

Factor, then cancel. Substituting 1 gives \({(1 + 1 - 2)/(1 - 1) = 0/0}\): a signal. Top and bottom are both 0 at 1, so both contain the factor \({x - 1}\):

\[ \begin{aligned} &\frac{x^2 + x - 2}{x^2 - 1} \\ &\quad= \frac{(x - 1)(x + 2)}{(x - 1)(x + 1)} \\ &\quad= \frac{x + 2}{x + 1} \qquad (x \neq 1). \end{aligned} \]

Cancelling is allowed because a limit never looks at \({x = 1}\). The new bottom is 2 at 1, not 0, so substitution now works: the limit is \({3/2}\).

Rationalise, then cancel. Substituting 9 gives \({(3 - 3)/(9 - 9) = 0/0}\) again. A square root minus a number is cleared by multiplying top and bottom by the same root plus the number, since \({(\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9}\):

\[ \begin{aligned} &\frac{\sqrt{x} - 3}{x - 9} \\ &\quad= \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)} \\ &\quad= \frac{x - 9}{(x - 9)(\sqrt{x} + 3)} \\ &\quad= \frac{1}{\sqrt{x} + 3} \qquad (x \neq 9). \end{aligned} \]

By the root, sum and quotient laws the limit is \({1/(3 + 3) = 1/6}\).

Two columns of four boxes joined by green downward arrows. Left column, titled Factor, then cancel: a pink box, x squared plus x minus 2 over x squared minus 1 at x equals 1 gives 0 over 0; an arrow labelled factor; a blue box, x minus 1 times x plus 2, over x minus 1 times x plus 1; an arrow labelled cancel x minus 1; a blue box, x plus 2 over x plus 1, for x not equal to 1; an arrow labelled put x equals 1; a yellow box, three halves. Right column, titled Rationalise, then cancel: a pink box, root x minus 3 over x minus 9 at x equals 9 gives 0 over 0; an arrow labelled times root x plus 3 over root x plus 3; a blue box, x minus 9 over x minus 9 times root x plus 3; an arrow labelled cancel x minus 9; a blue box, 1 over root x plus 3, for x not equal to 9; an arrow labelled put x equals 9; a yellow box, one sixth.
M04.1.2-F4

The two 0/0 limits taken apart. Each needs one move that exposes the common factor, one cancel, and one substitution.

Tables agree: at \({x = 0.99}\) and \({1.01}\) the first ratio is 1.502513 and 1.497512, closing on 1.5, and at \({x = 8.99}\) and \({9.01}\) the second is 0.166713 and 0.166620, closing on \({1/6 \approx 0.166667}\).

Code

Each hand result is checked against a table. The helper prints values just left and just right of the point.

import math

def table(f, a, steps=(0.1, 0.01, 0.001)):
    """Values of f at a - h and a + h for each step h."""
    for h in steps:
        print(f"   h = {h:<6} f(a - h) = {f(a - h):.6f}   f(a + h) = {f(a + h):.6f}")

def factor_case(x):
    return (x**2 + x - 2) / (x**2 - 1)

def root_case(x):
    return (math.sqrt(x) - 3) / (x - 9)

print("(x^2 + x - 2)/(x^2 - 1) near 1; by hand 3/2 = 1.5")
table(factor_case, 1)
print("(sqrt(x) - 3)/(x - 9) near 9; by hand 1/6 = 0.166667")
table(root_case, 9)
(x^2 + x - 2)/(x^2 - 1) near 1; by hand 3/2 = 1.5
   h = 0.1    f(a - h) = 1.526316   f(a + h) = 1.476190
   h = 0.01   f(a - h) = 1.502513   f(a + h) = 1.497512
   h = 0.001  f(a - h) = 1.500250   f(a + h) = 1.499750
(sqrt(x) - 3)/(x - 9) near 9; by hand 1/6 = 0.166667
   h = 0.1    f(a - h) = 0.167132   f(a + h) = 0.166206
   h = 0.01   f(a - h) = 0.166713   f(a + h) = 0.166620
   h = 0.001  f(a - h) = 0.166671   f(a + h) = 0.166662

Now the squeeze. At each \(x\) the ratio \({\sin(x)/x}\) should sit between \({\cos x}\) and 1.

for x in (0.5, 0.1, 0.01, 0.001):
    ratio = math.sin(x) / x
    print(f"x = {x:<6} cos x = {math.cos(x):.8f} <= sin(x)/x = {ratio:.8f} <= 1")
x = 0.5    cos x = 0.87758256 <= sin(x)/x = 0.95885108 <= 1
x = 0.1    cos x = 0.99500417 <= sin(x)/x = 0.99833417 <= 1
x = 0.01   cos x = 0.99995000 <= sin(x)/x = 0.99998333 <= 1
x = 0.001  cos x = 0.99999950 <= sin(x)/x = 0.99999983 <= 1

NumPy checks the squeeze at ten thousand points at once. It also shows the trap of the next section: with \(x\) in degrees, the same ratio heads to \({\pi / 180}\), not 1.

import numpy as np

xs = np.linspace(-1.5, 1.5, 10_000)       # an even number of points, so 0 is not one of them
ratio = np.sin(xs) / xs                   # (the ratio has no value at 0 itself)
print("cos x <= sin(x)/x everywhere:", bool(np.all(np.cos(xs) <= ratio)))
print("sin(x)/x <= 1 everywhere:    ", bool(np.all(ratio <= 1)))

degrees = np.array([1.0, 0.1, 0.01])
print("x in degrees:", np.sin(np.radians(degrees)) / degrees, " pi/180 =", np.pi / 180)
cos x <= sin(x)/x everywhere: True
sin(x)/x <= 1 everywhere:     True
x in degrees: [0.01745241 0.01745328 0.01745329]  pi/180 = 0.017453292519943295

Explore

This module’s interactive, the limit window, lives in lesson M04.1.1. Choose the curve \({y = x^2}\) and press Find δ for a few tolerances. The read-out gives the largest window as \({\sqrt{1 + \varepsilon} - 1}\), from solving \({(1 + \delta)^2 - 1 = \varepsilon}\), and the slider moves to it, rounded down to its steps of 0.01: algebra again, doing the work of the game.

Pitfalls

Two graphs on the same vertical scale from 0 to about 1.3. Left, marked with a cross and titled Degrees: not 1: the ratio sine x over x with x in degrees, from minus 90 to 90, is a blue curve lying almost flat along the bottom, with a hollow circle at x equals 0 just above zero and a pointer saying it heads to pi over 180, about 0.0175; a green dashed line at height 1 is labelled where it was expected: 1. Right, marked with a tick and titled Radians: the limit is 1: the same ratio with x in radians, from minus 1.6 to 1.6, is a blue arch with a hollow circle at its top, at (0, 1), labelled heads to 1.
M04.1.2-F5

Wrong beside right. Left: with \(x\) in degrees, \({\sin(x)/x}\) heads to \({\pi/180 \approx 0.0175}\). Right: with \(x\) in radians it heads to 1. The sector area \({x/2}\) is true only in radians.

“\({\sin(x)/x}\) tends to 1, whatever the units.”

The proof used the sector area \({x/2}\), which holds only when \(x\) is in radians. In degrees the sector is \({\tfrac{x}{360} \times \pi}\), and the ratio tends to \({\pi/180}\). Code works in radians too: in Python, math.sin(90) is the sine of 90 radians, not of a right angle.

“\({0/0}\) means the limit is 0” (or 1, or does not exist).

\({0/0}\) is not a value; it says only that substitution has failed. The three limits of this lesson that start as \({0/0}\) come out as \({3/2}\), \({1/6}\) and 1. Such a form is called indeterminate: the answer depends on how fast the top and the bottom shrink.

“Cancelling \({x - 1}\) changes the function, so it changes the limit.”

The cancelled form differs from the original at \({x = 1}\) only, where the original has no value. A limit never looks there, so both have the same limit.

“The quotient law still works when the bottom tends to 0.”

It needs \({m \neq 0}\). When the bottom tends to 0 and the top does not, the quotient grows without bound, as \({1/x}\) does near 0; the next lesson deals with that case.

Check yourself

Recall

1. State the sum law for limits.

Show answer

If \({f(x) \to \ell}\) and \({g(x) \to m}\) as \({x \to a}\), then \({f(x) + g(x) \to \ell + m}\).

2. When can the limit of a rational function \({p(x)/q(x)}\) at \(a\) be found by substituting \(a\)?

Show answer

When \({q(a) \neq 0}\). Then the limit is \({p(a)/q(a)}\).

3. Substituting \(a\) gives \({0/0}\). What does that tell you?

Show answer

Only that substitution has failed. The limit may be any number, or may not exist; factor and cancel, rationalise, or squeeze to find out.

4. State the squeeze theorem in words.

Show answer

If a function is trapped between two others near \(a\), and the two outer functions have the same limit at \(a\), the trapped one has that limit too.

5. What is \({\lim_{x \to 0} \sin(x)/x}\), and in which units must \(x\) be?

Show answer

1, with \(x\) in radians. In degrees the limit is \({\pi/180}\).

Exercises

1. ★ Find \({\lim_{x \to 3} (x^2 - 2x - 3)/(x - 3)}\).

Show solution

Substituting gives \({(9 - 6 - 3)/0 = 0/0}\), so factor the top. Its zeros are 3 and \({-1}\), so

\[ \begin{aligned} &\frac{x^2 - 2x - 3}{x - 3} \\ &\quad= \frac{(x - 3)(x + 1)}{x - 3} \\ &\quad= x + 1 \qquad (x \neq 3), \end{aligned} \]

and the limit is \({3 + 1 = 4}\). A table agrees: 3.9 at \({x = 2.9}\) and 4.1 at \({x = 3.1}\).

2. ★★ Find \({\lim_{x \to 0} (\sqrt{1 + x} - 1)/x}\).

Show solution

Substituting gives \({0/0}\). Multiply top and bottom by \({\sqrt{1 + x} + 1}\). On top this gives

\[ \begin{aligned} &(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1) \\ &\quad= (1 + x) - 1 = x , \end{aligned} \]

so

\[ \begin{aligned} &\frac{\sqrt{1 + x} - 1}{x} \\ &\quad= \frac{x}{x\,(\sqrt{1 + x} + 1)} \\ &\quad= \frac{1}{\sqrt{1 + x} + 1} \quad (x \neq 0). \end{aligned} \]

By the root, sum and quotient laws the limit is \({1/(1 + 1) = 1/2}\). At \({x = 0.01}\) the original is about 0.498756, and at \({x = -0.01}\) about 0.501256.

3. ★★★ Use \({\lim_{x \to 0} \sin(x)/x = 1}\) and the limit laws to find \({\lim_{x \to 0} (1 - \cos x)/x^2}\).

Show solution

Substituting gives \({0/0}\). Multiply top and bottom by \({1 + \cos x}\), and use \({1 - \cos^2 x = \sin^2 x}\):

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

As \(x\) tends to 0, \({\sin(x)/x}\) tends to 1, so its square tends to 1 by the power law, and \({\cos x}\) tends to 1, so \({1/(1 + \cos x)}\) tends to \({1/2}\) by the quotient law. By the product law the limit is \({1 \times \tfrac12 = \tfrac12}\). At \({x = 0.1}\) the ratio is about 0.499583.

Summary card

Limits follow the arithmetic of their parts; 0/0 is a signal to dig.

The anchor figure again: left, a blue curve with a hole at (1, 1.5), where cancelling a hidden factor reveals the limit three halves; right, sine x over x trapped in a shaded band between cosine x and 1, with a hole at (0, 1) where both bounds meet.
M04.1.2-F1

Cancel the factor, or squeeze.

  • Laws. Limits of sums, products and quotients (bottom limit not 0) are the sums, products and quotients of the limits.
  • Substitution. For polynomials, and for rational functions whose bottom is not 0 at \(a\).
  • \({0/0}\). Factor and cancel, or rationalise, then substitute.
  • Squeeze. \({\cos x \leq \sin(x)/x \leq 1}\), so \({\lim_{x \to 0} \sin(x)/x = 1}\), in radians.

Sources and next steps

Sources. Each was opened and checked against this lesson on 12 November 2026.

  • Gilbert Strang and Edwin Herman, Calculus Volume 1 (OpenStax), Section 2.3, The limit laws: the basic limits of \(x\) and of a constant (Theorem 2.4), the limit laws (Theorem 2.5), direct substitution for polynomials and rational functions (Theorem 2.6), the squeeze theorem (Theorem 2.7) and \({\lim_{\theta \to 0} \sin(\theta)/\theta = 1}\) (equation 2.18), which the book derives by comparing lengths on the unit circle where this lesson compares areas. Section 2.5 proves the sum law from the definition, under “Proving limit laws”.
  • Evan Dummit, Calculus I (part 2s): Trigonometric Limits, Northeastern University, 2016, Section 2.8: the proof that \({\sin(x)/x}\) tends to 1 from the areas \({\tfrac12 \sin x}\), \({\tfrac12 x}\) and \({\tfrac12 \tan x}\) and the squeeze theorem, and the limit \({(1 - \cos x)/x^2 \to \tfrac12}\) of Exercise 3.

Next. M04.1.3 One-sided limits and infinity lets \(x\) approach from one side only, and lets either \(x\) or the output run off to infinity, where the sigmoid levels off.

This lesson unlocks

M04.1.3 One-sided limits and infinity

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