Math · Calculus I · Worked example

A trigonometric limit using sin u / u → 1

Use the special sine limit to evaluate this expression. Angles are in radians.

lim_{θ \to 0} \frac{1 - cos θ}{sin (21 θ)}

Try direct substitution first

At $θ = 0$ the numerator and the denominator are both $0$ . Substitution gives $0 / 0$ , the same signal as in the factoring example. Here there is no polynomial factor to cancel, so the rewrite has to come from trigonometry.

\frac{1 - cos 0}{sin (21 \cdot 0)} = \frac{0}{0}

Why isn’t

0 / 0

the answer?

Zero divided by zero is undefined. Here “ $0 / 0$ ” describes what substitution produces, not the value of the limit. Different expressions with this form approach different values.

Turn the numerator into sines

The half-angle identity replaces $1 - cos θ$ with a squared sine. Both the numerator and the denominator are then sines, which is the shape the special limit needs.

\begin{array}{c} 1 - cos θ = 2 {sin}^{2} (\frac{θ}{2}) \\ lim_{θ \to 0} \frac{2 {sin}^{2} (\frac{θ}{2})}{sin (21 θ)} \end{array}

Aim for $sin u / u$

The goal is to build quotients of the form $sin u / u$ , because each one approaches $1$ as $u$ approaches $0$ . For $u = 0.01$ , $sin u / u \approx 0.99998$ already. Building one means multiplying and dividing by the same quantity, which leaves the value unchanged. The angle inside each sine and the quantity divided under it must match.

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

Which angles should match?

For $sin (θ / 2)$ , use $u = θ / 2$ . For $sin (21 θ)$ , use $u = 21 θ$ . Both angles approach $0$ as $θ$ approaches $0$ . The special sine limit holds in radians.

\begin{array}{c} u = \frac{θ}{2} \to 0 \\ u = 21 θ \to 0 \end{array}

Build the matching factors

Divide each copy of $sin (θ / 2)$ by $θ / 2$ , and multiply by ${(θ / 2)}^{2}$ to make up for it. Then multiply and divide by $21 θ$ . That makes the reciprocal ratio on the right and leaves an ordinary algebraic factor in the middle.

lim_{θ \to 0} 2 {(\frac{sin (\frac{θ}{2})}{\frac{θ}{2}})}^{2} \frac{{(\frac{θ}{2})}^{2}}{21 θ} (\frac{21 θ}{sin (21 θ)})

\begin{array}{c} lim_{θ \to 0} 2 {(\frac{sin (\frac{θ}{2})}{\frac{θ}{2}})}^{2} \\ \cdot \frac{{(\frac{θ}{2})}^{2}}{21 θ} \\ \cdot (\frac{21 θ}{sin (21 θ)}) \end{array}

Why are the added factors allowed?

Each inserted pair multiplies to $1$ . The rewrites hold near $0$ with $θ \neq 0$ , and they make no claim that the original expression is defined at $0$ . Taking $0 < ∣ θ ∣ < π / 21$ also keeps $sin (21 θ)$ away from $0$ .

\begin{array}{c} \frac{{(\frac{θ}{2})}^{2}}{{(\frac{θ}{2})}^{2}} = 1 \\ \frac{21 θ}{21 θ} = 1 \end{array}

Simplify the middle factor

Set the trigonometry aside for a moment. Square $θ / 2$ , combine the constants $4$ and $21$ , and cancel one factor of $θ$ . Canceling is allowed because $θ \neq 0$ on the way to the limit.

\begin{aligned} \frac{{(\frac{θ}{2})}^{2}}{21 θ} & = \frac{θ^{2} / 4}{21 θ} \\ = \frac{θ^{2}}{84 θ} \\ = \frac{θ}{84} \end{aligned}

Put the simpler factor back

Only the middle factor has changed. The two sine ratios still match the special limit, and $θ / 84$ is the part that will approach $0$ .

lim_{θ \to 0} 2 {(\frac{sin (\frac{θ}{2})}{\frac{θ}{2}})}^{2} (\frac{θ}{84}) (\frac{21 θ}{sin (21 θ)})

\begin{array}{c} lim_{θ \to 0} 2 {(\frac{sin (\frac{θ}{2})}{\frac{θ}{2}})}^{2} \\ \cdot \frac{θ}{84} \\ \cdot (\frac{21 θ}{sin (21 θ)}) \end{array}

Find each factor’s limit

The left ratio approaches $1$ , and the middle factor approaches $0$ . On the right, $sin (21 θ) / (21 θ)$ approaches $1$ , so its reciprocal approaches $1$ too. Taking the reciprocal is safe because that limit is not $0$ .

\begin{array}{c} \frac{sin (\frac{θ}{2})}{\frac{θ}{2}} \to 1 \\ \frac{θ}{84} \to 0 \\ \frac{21 θ}{sin (21 θ)} \to 1 \end{array}

Why does the reciprocal also approach

1

The quotient law for limits allows a reciprocal when the limit underneath is not $0$ . The ratio $sin (21 θ) / (21 θ)$ approaches $1$ , so it stays away from $0$ close enough to the limit.

\frac{21 θ}{sin (21 θ)} = \frac{1}{sin (21 θ) / (21 θ)} \to 1

\begin{array}{c} \frac{21 θ}{sin (21 θ)} \\ = \frac{1}{sin (21 θ) / (21 θ)} \to 1 \end{array}

Multiply the limits

All three factors have finite limits, so the product law applies. Keep the coefficient $2$ and the square on the first ratio. The $0$ in the middle makes the whole product $0$ .

2 {(1)}^{2} (0) (1) = 0

Why the answer makes sense

Near $0$ the numerator shrinks roughly like $θ^{2} / 2$ , while the denominator shrinks roughly like $21 θ$ . One factor of $θ$ is left over after canceling, and the quotient approaches $0$ from either side. These approximations explain the result, and the exact factorization above establishes it.

\begin{array}{c} 1 - cos θ \approx \frac{θ^{2}}{2} \\ sin (21 θ) \approx 21 θ \\ \frac{θ^{2} / 2}{21 θ} = \frac{θ}{42} \to 0 \end{array}

Result

The limit is $0$ . The expression is undefined at $θ = 0$ , but its nearby values approach $0$ .

0

Your turn

1. One like the example

Replace $21$ with $7$ in the denominator. What is the limit of $(1 - cos θ) / sin (7 θ)$ as $θ$ approaches $0$ ?

Show the answer and explanation

The limit is still $0$ . Near $0$ the quotient behaves like $θ / 14$ .

The same factorization works with $7 θ$ in place of $21 θ$ . The middle factor becomes ${(θ / 2)}^{2} / (7 θ) = θ / 28$ , which approaches $0$ , and the two sine ratios approach $1$ . As a check, the numerator is about $θ^{2} / 2$ and $sin (7 θ)$ is about $7 θ$ , so the quotient is about $θ / 14$ .

\begin{array}{l} lim_{θ \to 0} \frac{1 - cos θ}{sin (7 θ)} \\ 0 \end{array}

2. A variation

Find the limit of $sin (5 x) / (3 x)$ as $x$ approaches $0$ . Here the angle and the denominator do not match.

Show the answer and explanation

The limit is $5 / 3$ .

The special limit needs the angle $5 x$ under the sine. Multiply and divide by $5$ to write $sin (5 x) / (3 x)$ $= (5 / 3) \cdot sin (5 x) / (5 x)$ . The ratio approaches $1$ , so the limit is $5 / 3$ .

\begin{array}{l} lim_{x \to 0} \frac{sin (5 x)}{3 x} \\ lim_{x \to 0} \frac{5}{3} \cdot \frac{sin (5 x)}{5 x} \\ \frac{5}{3} \end{array}

3. Somewhere new

On a circle of radius $1$ , an arc of angle $θ$ has length $θ$ , and the straight chord across it has length $2 sin (θ / 2)$ . What does $chord \div arc$ approach as $θ$ approaches $0$ ?

Show the answer and explanation

It approaches $1$ . A short arc is almost straight.

The ratio is $2 sin (θ / 2) / θ$ , which is $sin (θ / 2) / (θ / 2)$ . With $u = θ / 2$ , this is $sin u / u$ , and $u$ approaches $0$ with $θ$ . Its limit is $1$ , so a tiny arc and its chord have nearly the same length.

\begin{array}{l} lim_{θ \to 0} \frac{2 sin (\frac{θ}{2})}{θ} \\ lim_{θ \to 0} \frac{sin (\frac{θ}{2})}{\frac{θ}{2}} \\ 1 \end{array}

Keep exploring

Open the steps in Math: the Checker marks the identity, the factored product and the value $0$ ✓. Change the $0$ to $1 / 42$ and its mark turns to ✗.

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

OpenStax Calculus Volume 1 — The Limit Laws (special sine limit and squeeze theorem)

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 5 key terms Check the limit in Math Open worked example on a board Calculus formulas

Your existing work stays on this device. Examples open as editable copies.

Try direct substitution first

Turn the numerator into sines

Aim for sinu⁢/⁢usinu/u

Build the matching factors

Simplify the middle factor

Put the simpler factor back

Find each factor’s limit

Multiply the limits

Why the answer makes sense

Result

Your turn

1. One like the example

The limit is still 00. Near 00 the quotient behaves like θ⁢/14θ/14.

2. A variation

The limit is 5⁢/35/3.

3. Somewhere new

It approaches 11. A short arc is almost straight.

Keep exploring

Study it, then try it

Aim for $sin u / u$

The limit is still $0$ . Near $0$ the quotient behaves like $θ / 14$ .

The limit is $5 / 3$ .

It approaches $1$ . A short arc is almost straight.