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Limit Laws and Algebraic Methods · Tutorial 214 of 1495

Trigonometric Identity Rewrites for Limits

Use trigonometric identities to transform indeterminate quotients into expressions evaluated with standard sine and cosine limits.

Beginner 10 min read

What You'll Learn

  • Recognize when direct substitution gives a zero-over-zero trigonometric quotient
  • Use the half-angle identity for one minus cosine
  • Apply double-angle and Pythagorean identities to simplify limit expressions
  • Reduce scaled sine quotients to the standard sine limit
  • Track when cancellation is valid in a punctured neighborhood
  • Check rewritten expressions and identify a common half-angle error

Why Rewrite a Trigonometric Quotient?

When direct substitution into a trigonometric quotient gives \(0/0\), the expression does not by itself determine the limit. A useful next step is to rewrite it with a trigonometric identity. The goal is to expose a standard limit, especially \(\lim_{u\to 0}\frac{\sin u}{u}=1\), established in the earlier material on the sine quotient limit.

This tutorial focuses on identities that turn expressions involving \(1-\cos u\), \(\sin(2u)\), and squared trigonometric functions into forms that can be evaluated using that standard limit and the continuity of cosine. Angles are measured in radians throughout; the standard sine quotient limit in this form requires radians.

Strategy: Identify the trigonometric pattern, use an identity to rewrite it, and then separate the expression into standard sine quotients and factors with known limits. Only cancel a factor when the cancellation is valid for nearby inputs where the original quotient is defined.

Identities Used in Limit Rewrites

The half-angle identity is particularly useful when a numerator contains one minus cosine. It follows from the double-angle identity \(\cos(2t)=1-2\sin^2 t\): setting \(t=u/2\) gives the equivalent formula below.

$$ 1-\cos u=2\sin^2\left(\frac{u}{2}\right). $$

Other useful identities are the double-angle formula for sine and the Pythagorean identity. Each is valid for every real \(u\), so it may be substituted into an expression wherever that expression is defined.

$$ \sin(2u)=2\sin u\cos u, \qquad \sin^2 u+\cos^2 u=1, \qquad 1-\cos^2 u=\sin^2 u. $$

To see why the half-angle identity is well suited to limits near zero, put it into the quotient with \(u^2\) in the denominator. For \(u\ne 0\), the sine quotient can be isolated:

$$ \frac{1-\cos u}{u^2} = \frac{2\sin^2(u/2)}{u^2} = \frac{1}{2}\left(\frac{\sin(u/2)}{u/2}\right)^2. $$

As \(u\to 0\), \(u/2\to 0\), so the standard sine quotient limit applies. Squaring and multiplying by \(1/2\) then gives the quadratic cosine limit.

Result: Using the half-angle identity and the standard sine quotient limit, \(\lim_{u\to 0}\frac{1-\cos u}{u^2}=\frac12\). The quotient identity used to obtain this result holds for \(u\ne 0\); that is sufficient for a limit at zero.

Worked Example: A One-Minus-Cosine Quotient

Worked Example: A One-Minus-Cosine Quotient

Evaluate \(\displaystyle\lim_{x\to 0}\frac{1-\cos(4x)}{x^2}\).

Direct substitution gives \((1-1)/0^2=0/0\), so substitution alone does not evaluate the limit. Apply the half-angle identity with \(u=4x\):

$$ \begin{aligned} \frac{1-\cos(4x)}{x^2} &=\frac{2\sin^2(2x)}{x^2}\\ &=8\left(\frac{\sin(2x)}{2x}\right)^2. \end{aligned} $$

For the second equality, \(8(\sin(2x)/(2x))^2=8\sin^2(2x)/(4x^2)=2\sin^2(2x)/x^2\), as required. Since \(2x\to 0\), the standard sine quotient tends to \(1\). Therefore,

$$ \lim_{x\to 0}\frac{1-\cos(4x)}{x^2} =8(1)^2=8. $$

Check the coefficient by expressing the quotient in terms of the quadratic cosine limit: for \(x\ne0\), it equals \(16\frac{1-\cos(4x)}{(4x)^2}\). The latter has limit \(16\cdot\frac12=8\), agreeing with the identity rewrite.

Scaled Sine Quotients After Rewriting

Many problems reduce to a ratio of sine functions whose arguments are constant multiples of the variable. If \(a\) and \(b\) are nonzero constants, write each sine in terms of its own argument divided by that argument. For \(x\ne0\) sufficiently near zero,

$$ \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b} \frac{\sin(ax)}{ax} \frac{bx}{\sin(bx)}. $$

As \(x\to0\), both \(ax\) and \(bx\) tend to zero. The first sine quotient tends to \(1\), and the reciprocal of the second tends to \(1\), since \(\sin(bx)/(bx)\to1\). Thus the ratio tends to \(a/b\). This result is useful after applying an identity, but its conditions matter: both scale factors must be nonzero, and the quotient is considered near zero where its denominator is defined.

Limit rule: For nonzero constants \(a\) and \(b\), \(\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b}\). This follows by factoring out the standard sine quotients; it is not a rule for substituting into a quotient that is still \(0/0\).

Worked Example: A Squared-Sine Quotient

Worked Example: A Squared-Sine Quotient

Evaluate \(\displaystyle\lim_{x\to0}\frac{1-\cos^2(3x)}{\sin^2(2x)}\).

At \(x=0\), both the numerator and denominator are zero. Use \(1-\cos^2 u=\sin^2 u\), with \(u=3x\):

$$ \frac{1-\cos^2(3x)}{\sin^2(2x)} = \frac{\sin^2(3x)}{\sin^2(2x)} = \left(\frac{\sin(3x)}{\sin(2x)}\right)^2. $$

For \(x\ne0\) sufficiently near zero, apply the scaled sine quotient result with \(a=3\) and \(b=2\). The inner quotient tends to \(3/2\), so the original expression tends to its square:

$$ \lim_{x\to0}\left(\frac{\sin(3x)}{\sin(2x)}\right)^2 = \left(\frac32\right)^2 = \frac94. $$

Check the rewrite by squaring the ratio: \((\sin(3x)/\sin(2x))^2=\sin^2(3x)/\sin^2(2x)\), exactly the expression obtained from the Pythagorean identity. The limit is therefore \(\frac94\).

Worked Example: Combining a Double-Angle Identity with a Cosine Factor

Worked Example: Combining a Double-Angle Identity with a Cosine Factor

Evaluate \(\displaystyle\lim_{x\to0}\frac{\sin(2x)}{x(1+\cos x)}\).

At zero, the numerator is \(0\) and the denominator is \(0\). Use \(\sin(2x)=2\sin x\cos x\), then rearrange for \(x\ne0\):

$$ \begin{aligned} \frac{\sin(2x)}{x(1+\cos x)} &=\frac{2\sin x\cos x}{x(1+\cos x)}\\ &=2\left(\frac{\sin x}{x}\right)\frac{\cos x}{1+\cos x}. \end{aligned} $$

The first factor in parentheses tends to \(1\). By continuity of cosine, \(\cos x\to1\), so the remaining fraction tends to \(1/(1+1)=1/2\); its denominator is nonzero near zero. Hence,

$$ \lim_{x\to0}\frac{\sin(2x)}{x(1+\cos x)} =2\cdot1\cdot\frac12=1. $$

Check the coefficient using the double-angle identity in the original numerator: after substitution the quotient is exactly \(2(\sin x/x)(\cos x/(1+\cos x))\). Its limiting factors multiply to \(2\cdot1\cdot\frac12=1\), confirming the result.

Cancellation and a Common Mistake

Rewriting identities does not make cancellation automatically valid at the point of the limit. For example, in the first worked example the factor \(x\) in \(\sin(2x)/(2x)\) is nonzero when \(x\ne0\). A limit as \(x\to0\) concerns nearby inputs, not necessarily the value at \(x=0\). It is therefore legitimate to use an equality that holds for all sufficiently close nonzero \(x\), even if the original expression is undefined at zero.

A frequent error is to confuse the half-angle identity with \(1-\cos x=2\sin^2 x\). That formula is incorrect: the correct sine argument is \(x/2\). For a concrete check, let \(x=\pi/3\). Then \(1-\cos(\pi/3)=1-\frac12=\frac12\), whereas \(2\sin^2(\pi/3)=2(\frac{\sqrt3}{2})^2=\frac32\). The incorrect identity would also give a wrong limit: applying it to \((1-\cos x)/x^2\) would produce \(2(\sin x/x)^2\), whose limit is \(2\), rather than the correct value \(\frac12\).

Check the argument: The identity is \(1-\cos u=2\sin^2(u/2)\). When the cosine input is \(4x\), for example, the sine input must be \(2x\). Keeping track of the halved argument prevents a wrong coefficient in the limit.

A Practical Rewrite Routine

When a trigonometric quotient gives \(0/0\) at a point, use a deliberate sequence rather than trying identities at random.

1
Locate the vanishing parts.
Substitute the limiting input to see whether the numerator and denominator both approach zero.
2
Match a pattern to an identity.
Use \(1-\cos u=2\sin^2(u/2)\), \(\sin(2u)=2\sin u\cos u\), or \(1-\cos^2 u=\sin^2 u\) when the expression contains the matching terms.
3
Expose standard limits.
Rewrite sine factors using \(\sin v/v\), with \(v\to0\), and keep all scale constants visible.
4
Evaluate the remaining factors.
Use the standard sine quotient limit and continuity of cosine, checking that any denominator left in a factor has a nonzero limit.
5
Verify the rewrite.
Check the identity's input and coefficient, and confirm that any cancellation was made only where the canceled factor is nonzero.

These rewrites matter because a quotient that initially appears indeterminate can conceal a straightforward product of known limiting factors. The identity supplies the algebraic bridge; the standard limits supply the values. A reliable solution shows both, so that its coefficients, arguments, and conditions can all be checked.

Check Your Understanding

For each question, pay attention to the identity used and to the argument of each sine quotient.

  1. Use the half-angle identity to rewrite \(\frac{1-\cos(2x)}{x^2}\) in terms of \(\sin x/x\), then evaluate its limit as \(x\to0\).
  2. Evaluate \(\displaystyle\lim_{x\to0}\frac{1-\cos(6x)}{x^2}\) by rewriting the numerator.
  3. Use \(1-\cos^2 u=\sin^2 u\) to evaluate \(\displaystyle\lim_{x\to0}\frac{1-\cos^2(5x)}{\sin^2(2x)}\).
  4. State the condition on the constants \(a\) and \(b\) in the scaled sine quotient limit \(\lim_{x\to0}\sin(ax)/\sin(bx)=a/b\).
  5. What is the correct identity for \(1-\cos u\), and what specific error occurs in the incorrect formula \(1-\cos u=2\sin^2 u\)?

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