Tutorials › Calculus I › Deriving the Quadratic Cosine Limit

Limit Laws and Algebraic Methods · Tutorial 212 of 1495

Deriving the Quadratic Cosine Limit

Use the half-angle identity and the sine quotient limit to derive the quadratic cosine limit and evaluate related limits with scaled angles.

Beginner 9 min read

What You'll Learn

  • State the quadratic cosine limit and its radian-measure condition.
  • Rewrite one minus cosine using a half-angle identity.
  • Track the scale factor when the sine quotient has argument x divided by a constant.
  • Evaluate limits with scaled cosine arguments and constant denominator factors.
  • Identify common errors involving zero-over-zero forms and missing scale factors.

Why a Cosine Difference Needs More Than Substitution

Direct substitution into \((1-\cos x)/x^2\) at \(x=0\) gives \(0/0\), so the quotient's limit is not determined by evaluating its numerator and denominator separately. The expression has a useful structure, however: the difference \(1-\cos x\) can be rewritten as a square involving sine. That rewrite connects the problem to the sine quotient limit established earlier.

The angle convention matters. In this tutorial, \(x\) is measured in radians. The standard sine quotient limit used below is \(\lim_{u\to 0}\sin u/u=1\) when \(u\) is in radians. Using that limit with a degree-valued angle would give a different numerical result.

Key result: When \(x\) is measured in radians, \(\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac12.\)

The Identity That Reveals the Limit

The double-angle identity for cosine is \(\cos x=1-2\sin^2(x/2)\). Rearranging it gives \(1-\cos x=2\sin^2(x/2)\). For \(x\ne 0\), divide this identity by \(x^2\). Since \(x^2=4(x/2)^2\), the quotient becomes

$$ \frac{1-\cos x}{x^2} = \frac{2\sin^2(x/2)}{4(x/2)^2} = \frac12\left(\frac{\sin(x/2)}{x/2}\right)^2. $$

This is an equality for every nonzero \(x\), not just an approximation near zero. The expression in parentheses is a sine quotient whose argument \(x/2\) approaches zero as \(x\) does. The sine quotient limit therefore determines the limit of the whole expression.

Definition: The quadratic cosine limit is the limit of \((1-\cos x)/x^2\) as \(x\) approaches zero. Its value is \(1/2\) when \(x\) is measured in radians.

Derivation of the Quadratic Cosine Limit

Here is the complete limit argument. The sine quotient limit says that \(\sin u/u\) approaches \(1\) as \(u\) approaches zero in radians. Set \(u=x/2\). Then \(x\to0\) implies \(u\to0\), so \(\sin(x/2)/(x/2)\to1\). The limit laws for products and powers allow us to square this limit and multiply by \(1/2\).

1
Rewrite the cosine difference.
For every \(x\ne0\), use \(1-\cos x=2\sin^2(x/2)\) and \(x^2=4(x/2)^2\) to obtain \(\displaystyle \frac{1-\cos x}{x^2}=\frac12\left(\frac{\sin(x/2)}{x/2}\right)^2.\)
2
Apply the sine quotient limit.
As \(x\to0\), \(x/2\to0\), and hence \(\displaystyle \frac{\sin(x/2)}{x/2}\to1.\)
3
Take the limit of the rewritten expression.
The limit laws give \(\displaystyle \frac12\left(\frac{\sin(x/2)}{x/2}\right)^2\to\frac12(1)^2=\frac12.\)
$$ \boxed{\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12} $$

The equality used in Step 1 holds on a punctured neighbourhood of zero: it holds for every nonzero \(x\), even though the quotient itself is not defined at zero. Limits depend on the values arbitrarily close to the point, not on a value at the point itself. This is why the algebraic rewrite is sufficient.

Worked Examples

Worked Example: The Basic Quadratic Cosine Limit

Evaluate \(\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}\), with \(x\) in radians.

Direct substitution gives \(1-\cos 0=0\) and \(0^2=0\), so the quotient has the indeterminate form \(0/0\). Instead, use the identity:

$$ \begin{aligned} \frac{1-\cos x}{x^2} &=\frac{2\sin^2(x/2)}{x^2}\\ &=\frac{2\sin^2(x/2)}{4(x/2)^2}\\ &=\frac12\left(\frac{\sin(x/2)}{x/2}\right)^2. \end{aligned} $$

As \(x\to0\), \(x/2\to0\), so the sine quotient tends to \(1\). Therefore

$$ \lim_{x\to0}\frac{1-\cos x}{x^2} =\frac12(1)^2 =\frac12. $$

Check: The identity shows that the original quotient equals one half of the squared sine quotient for each \(x\ne0\). Since that sine quotient tends to \(1\), the resulting value is indeed one half.

Worked Example: A Scaled Angle

Evaluate \(\displaystyle\lim_{x\to0}\frac{1-\cos(3x)}{x^2}\), with \(x\) in radians.

Apply the half-angle identity to the angle \(3x\):

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

To use the sine quotient limit, form the quotient with the same argument in numerator and denominator:

$$ \begin{aligned} \frac{1-\cos(3x)}{x^2} &=\frac{2\sin^2(3x/2)}{x^2}\\ &=2\left(\frac{\sin(3x/2)}{3x/2}\right)^2 \left(\frac{3x/2}{x}\right)^2\\ &=2\left(\frac{\sin(3x/2)}{3x/2}\right)^2 \left(\frac32\right)^2\\ &=\frac92\left(\frac{\sin(3x/2)}{3x/2}\right)^2. \end{aligned} $$

As \(x\to0\), \(3x/2\to0\), so the sine quotient in the final line tends to \(1\). Thus

$$ \lim_{x\to0}\frac{1-\cos(3x)}{x^2} =\frac92(1)^2 =\frac92. $$

Check: The scale factor is squared: \((3/2)^2=9/4\), and the identity contributes the factor \(2\). Their product is \(2\cdot9/4=9/2\), matching the computed limit.

Worked Example: A Constant in the Denominator

Evaluate \(\displaystyle\lim_{x\to0}\frac{1-\cos(2x)}{5x^2}\), with \(x\) in radians.

First use the identity \(1-\cos(2x)=2\sin^2 x\). Then express the quotient using \(\sin x/x\):

$$ \begin{aligned} \frac{1-\cos(2x)}{5x^2} &=\frac{2\sin^2 x}{5x^2}\\ &=\frac25\left(\frac{\sin x}{x}\right)^2. \end{aligned} $$

As \(x\to0\), the sine quotient tends to \(1\). Therefore

$$ \lim_{x\to0}\frac{1-\cos(2x)}{5x^2} =\frac25(1)^2 =\frac25. $$

Check: The identity contributes the numerator factor \(2\), and the denominator contributes the factor \(5\). After the sine quotient tends to \(1\), the remaining constant is \(2/5\), as found.

Keeping Track of Scale Factors

For a fixed real constant \(a\ne0\), the same reasoning gives a useful consequence. The angle \(ax\) approaches zero with \(x\). For \(x\ne0\),

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

The sine quotient tends to \(1\), so the limit is \(a^2/2\). If the denominator is instead \(c x^2\), where \(c\ne0\), dividing by \(c\) gives \(a^2/(2c)\). These consequences rely on a fixed scale \(a\) and on the angle being measured in radians.

Scale-factor rule: For fixed \(a\in\mathbb R\) and \(c\ne0\), with \(x\) in radians, \(\displaystyle \lim_{x\to0}\frac{1-\cos(ax)}{c x^2}=\frac{a^2}{2c}.\) For \(a=0\), the numerator is identically zero, so the same formula also gives the correct value \(0\).

Why Radian Measure Is Essential

The sine quotient limit \(\sin u/u\to1\) is a statement about radian measure. Radians measure an angle by the ratio of arc length to radius. In this convention, the angle in radians is directly the ratio of those two lengths, which is the convention under which the standard sine quotient limit has value \(1\).

If a numerical variable \(d\) represents an angle in degrees, the corresponding radian measure is \(\pi d/180\). Consequently, the quadratic cosine limit with \(d\) in degrees is not \(1/2\). Substituting the radian angle gives

$$ \begin{aligned} \lim_{d\to0}\frac{1-\cos(\text{\(d\) degrees})}{d^2} &=\lim_{d\to0}\frac{1-\cos(\pi d/180)}{d^2}\\ &=\frac12\left(\frac{\pi}{180}\right)^2\\ &=\frac{\pi^2}{64800}. \end{aligned} $$

The factor \(\pi/180\) converts each degree to radians, and the conversion factor is squared because the denominator is quadratic. Thus the radian condition is part of the theorem, not an optional notation choice.

A Common Scale-Factor Mistake

A frequent error is to treat \(\sin(x/2)/x\) as though its limit were \(1\). The sine quotient limit applies to \(\sin u/u\), where the numerator's angle and denominator use the same argument. Replacing \(u\) by \(x/2\) gives \(\sin(x/2)/(x/2)\to1\), not \(\sin(x/2)/x\to1\).

$$ \text{Incorrect:}\qquad \frac{1-\cos x}{x^2} =\frac12\left(\frac{\sin(x/2)}{x}\right)^2 \ \Longrightarrow\ \frac12. $$

The displayed equality is incorrect: the identity gives \(2\sin^2(x/2)/x^2\), which is \(2\bigl(\sin(x/2)/x\bigr)^2\), not one half times that square. The correct conversion inserts \(x/2\) in the denominator of the sine quotient:

$$ \frac{1-\cos x}{x^2} =\frac12\left(\frac{\sin(x/2)}{x/2}\right)^2 \longrightarrow \frac12(1)^2 =\frac12. $$

Keeping the argument identical in the numerator and denominator makes the sine quotient limit applicable and prevents the scale factor from being lost.

When This Limit Is Useful

The quadratic cosine limit evaluates expressions where direct substitution leaves a cosine difference of the form \(1-\cos(\text{small angle})\) over a squared quantity. The half-angle identity converts that difference into a sine square; the sine quotient limit then supplies the finite value. In later calculations, the scale-factor rule lets you account for a constant multiplying the angle without repeating the full derivation.

The main checks are simple: confirm that the angle tends to zero, keep the angle in radians, and preserve any constant scale factors when matching the sine argument to its denominator. A missing square or an unmatched argument changes the limit.

Check Your Understanding

For each question, state the relevant identity or limit law before giving the result.

  1. Why does direct substitution into \((1-\cos x)/x^2\) fail to determine its limit?
  2. Which trigonometric identity rewrites \(1-\cos x\) as a square?
  3. Evaluate \(\displaystyle\lim_{x\to0}\frac{1-\cos(4x)}{x^2}\), with \(x\) in radians.
  4. Evaluate \(\displaystyle\lim_{x\to0}\frac{1-\cos x}{7x^2}\), with \(x\) in radians.
  5. Why does the value \(1/2\) depend on measuring \(x\) in radians?

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