Math · Calculus I · Worked example

Rewriting a trigonometric limit

Use the special sine limit to evaluate this expression. Angles are in radians; the denominator is the sine of twenty-one times θ.

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

Try direct substitution first

At θ = 0, the numerator and denominator both become zero. This gives the indeterminate form 0/0: substitution alone cannot tell us the limit. We need to rewrite the expression before we can evaluate it.

\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 can approach different values.

Use an identity to turn the numerator into sines

The half-angle identity replaces 1 − cos θ with a squared sine. That gives us sine expressions in both the numerator and denominator, which is useful for the special limit we want to apply.

\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}

Choose the form we want to create

We want to manufacture expressions of the form sin(u)/u, because their limits equal 1. “Manufacture” means multiplying and dividing by the same quantity so the value stays unchanged. The angle inside sine and the quantity beneath it must match.

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

Which angles should we match?

For sin(θ/2), use u = θ/2. For sin(21θ), use u = 21θ. Both angles approach zero as θ approaches zero. The standard sine limit uses 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, then multiply by (θ/2)² to compensate. Next, multiply and divide by 21θ. This creates the reciprocal sine 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}

Check why the added factors are allowed

Each inserted pair multiplies to 1. These rewrites apply near zero with θ ≠ 0; we are not claiming the original expression is defined at zero. Taking 0 < |θ| < π/21 also keeps sin(21θ) nonzero.

\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

Now set the trigonometry aside for a moment. Square θ/2, combine the constants 4 and 21, and cancel one factor of θ. That cancellation is valid for the nonzero values of θ used to approach 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 that simpler factor back

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

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}

Evaluate each factor’s limit

The left ratio approaches 1. The middle factor approaches 0. On the right, sin(21θ)/(21θ) approaches 1, so its reciprocal also approaches 1. Taking a reciprocal is safe here because the limiting value is nonzero.

\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 reciprocal limit law lets us take the reciprocal of a nonzero limiting value. The denominator ratio approaches 1, so it stays away from zero sufficiently close 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 we can use the product limit law. Keep the coefficient 2 and the square on the first ratio. The zero in the middle makes the product zero.

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

Why this makes sense

Near zero, the numerator shrinks roughly like θ², while the denominator shrinks roughly like θ. After cancellation, there is still one factor of θ left, so the quotient approaches zero from either side. These approximations explain the result; the exact factorization above established 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 zero.

0

Your turn

Replace 21 with 7 in the denominator. What is the new limit, and what simple expression describes its behavior near zero?

Show the answer and explanation

The limit is still 0; near zero the quotient behaves like θ/14.

The numerator is still approximately θ²/2, but sin(7θ) is approximately 7θ. Their quotient is approximately θ/14, which approaches zero. In the exact factorization, the middle factor becomes θ/28, and the leading 2 gives the same local behavior.

\begin{array}{c} lim_{θ \to 0} \frac{1 - cos θ}{sin (7 θ)} = 0 \\ \frac{θ^{2} / 2}{7 θ} = \frac{θ}{14} \end{array}

Keep exploring

When you meet another trigonometric 0/0 limit, look for a useful identity, then match each sine to its own angle.

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)