Why the Inner Function Matters
Expressions involving \(1-\cos u\) behave differently from ordinary linear differences near zero: the cosine defect becomes small at a rate proportional to \(u^2\). This quadratic behavior makes the ratio \(\frac{1-\cos u}{u^2}\) useful when a complicated inner function approaches zero.
The main task in a composite limit is to identify that inner function and then compare its square with the denominator in the original expression. A constant factor in the inner function becomes a squared factor in the limit, so it must not be discarded.
The Basic Cosine Defect Limit
All trigonometric limits here use radians. For a real variable \(u\) approaching zero, the quadratic cosine defect limit is
One way to see why this constant is \(\frac{1}{2}\) is to use the half-angle identity \(1-\cos u=2\sin^2(u/2)\). The standard sine limit \(\lim_{v\to0}\frac{\sin v}{v}=1\) then gives a direct derivation.
As \(u\to0\), the input \(u/2\) also approaches zero. Therefore \(\frac{\sin(u/2)}{u/2}\to1\). Squaring preserves that limit, so the final expression approaches \(\frac{1}{2}(1)^2=\frac{1}{2}\). This establishes the cosine defect limit.
Using a Composite Inner Function
Let \(g\) be a function such that \(g(x)\to0\) as \(x\to a\). If \(g(x)\ne0\) for all \(x\) sufficiently close to \(a\), with \(x\ne a\), the basic limit can be applied to the input \(u=g(x)\). The composition law for limits then gives
The condition \(g(x)\ne0\) nearby ensures that the displayed quotient is defined on a punctured neighborhood of \(a\). The essential limiting condition is \(g(x)\to0\): the inner function must approach zero so that the cosine defect limit applies.
To evaluate an expression with a different denominator, factor it into the standard quotient and a second factor. For example, if \(h(x)\ne0\) near \(a\), then
If \(g(x)\to0\), the standard quotient tends to \(\frac{1}{2}\). If, in addition, \(\frac{g(x)}{h(x)}\to L\) for a finite real number \(L\), the product law gives
This form makes the scaling explicit: the final constant depends on the limiting ratio of the inner function to the denominator’s base, and that ratio is squared.
Worked Examples
Worked Example: A Composite Input with Its Own Square in the Denominator
Evaluate \(\lim_{x\to2}\frac{1-\cos(x^2-4)}{(x^2-4)^2}\).
Set \(g(x)=x^2-4\). At \(x=2\), the polynomial gives \(g(2)=2^2-4=4-4=0\), so \(g(x)\to0\) as \(x\to2\). Also,
For \(x\ne2\) sufficiently close to \(2\), \(x-2\ne0\) and \(x+2\ne0\), so \(g(x)\ne0\). The expression is exactly the standard composite quotient with \(g(x)\) in both the cosine input and the denominator:
As a check, the half-angle identity rewrites the quotient as \(\frac{1}{2}\left(\frac{\sin((x^2-4)/2)}{(x^2-4)/2}\right)^2\). Its sine ratio tends to \(1\), so the limit is again \(\frac{1}{2}(1)^2=\frac{1}{2}\).
Worked Example: A Scaled Linear Input
Evaluate \(\lim_{x\to0}\frac{1-\cos(4x)}{x^2}\).
The inner function is \(g(x)=4x\), which approaches zero as \(x\to0\). The denominator is \(x^2\), not \((4x)^2\), so first insert the square of the inner function and compensate with a factor:
For \(x\ne0\), the first factor has the standard cosine defect limit, with \(u=4x\to0\). Thus,
Check the constant using the half-angle identity: \(\frac{1-\cos(4x)}{x^2} =\frac{1}{2}\left(\frac{\sin(2x)}{2x}\right)^2\cdot16\). The sine ratio tends to \(1\), giving \(\frac{1}{2}\cdot1^2\cdot16=8\).
Worked Example: A Nested Sine Input
Evaluate \(\lim_{x\to0}\frac{1-\cos(\sin(3x))}{x^2}\).
Take \(g(x)=\sin(3x)\). Since \(3x\to0\) and sine is continuous at zero, \(g(x)\to\sin(0)=0\). Factor using \(g(x)^2\):
The first factor tends to \(\frac{1}{2}\) by the composite cosine defect limit. For the second factor, use the standard sine limit and show the scale explicitly:
Therefore, the original limit is
The constant checks out: the nested input \(\sin(3x)\) is approximately \(3x\) near zero, so its ratio to \(x\) tends to \(3\), whose square multiplies the basic factor \(\frac{1}{2}\).
Worked Example: A Polynomial Input Compared with a Linear Denominator
Evaluate \(\lim_{x\to1}\frac{1-\cos((x-1)(x+2))}{(x-1)^2}\).
Let \(g(x)=(x-1)(x+2)\). As \(x\to1\), the first factor \(x-1\to0\) and the second factor \(x+2\to3\), so \(g(x)\to0\). For \(x\ne1\) close to \(1\), \(g(x)\ne0\). Split the quotient:
The first factor tends to \(\frac{1}{2}\). In the second factor, cancellation is valid for \(x\ne1\):
Hence the limit is
As a check, \(g(x)/(x-1)=x+2\) has limiting value \(3\), so \(g(x)^2/(x-1)^2\to9\). Multiplying that factor by the cosine defect limit \(\frac{1}{2}\) gives \(\frac{9}{2}\), as obtained.
A Reliable Process
For a composite cosine defect limit, identify the inner function before manipulating the expression. Then determine whether the denominator already contains its square. If not, calculate the limiting ratio between the inner function and the base of the denominator.
Write \(g(x)\) for the complete input to cosine, and verify that \(g(x)\to0\).
Make sure the relevant denominator is nonzero for \(x\ne a\) sufficiently close to the limit point.
Rewrite the expression using \(\frac{1-\cos(g(x))}{g(x)^2}\), whose limit is \(\frac{1}{2}\).
Evaluate the limit of the ratio between \(g(x)\) and the base of the denominator, then square it.
Common Mistake: Losing the Scale
A frequent error is to replace a composite input by the basic limit without accounting for its scale relative to the denominator. For example, in \(\frac{1-\cos(4x)}{x^2}\), one might incorrectly write
The basic limit gives \(\frac{1}{2}\) for the quotient with denominator \((4x)^2\), not \(x^2\). Since \((4x)^2=16x^2\), the correct factorization is
The correction is to keep the squared scale. More generally, if the inner function divided by the denominator’s base tends to \(L\), the resulting limit is \(\frac{1}{2}L^2\), not merely \(\frac{1}{2}\).
When This Limit Is Useful
The cosine defect limit turns a trigonometric expression into a comparison of ordinary functions near the limiting input. That is especially useful when the cosine input is itself a polynomial, a scaled variable, or another trigonometric expression. Rather than expanding the entire composite expression, isolate the universal cosine factor and study only how quickly its input approaches zero.
The method also clarifies when it cannot be applied directly. If the inner function does not approach zero, the quadratic cosine defect limit at zero does not describe the expression. If the denominator does not match the square of the inner function, the leftover ratio must be evaluated; assuming it is \(1\) without checking can change the answer.
Check Your Understanding
For each limit, identify the inner function and keep track of any scale factor. Assume the trigonometric inputs are in radians.
- Evaluate \(\lim_{x\to0}\frac{1-\cos(2x)}{x^2}\).
- Evaluate \(\lim_{x\to0}\frac{1-\cos(x^2)}{x^4}\).
- Evaluate \(\lim_{x\to0}\frac{1-\cos(\sin(2x))}{x^2}\).
- For an inner function \(g(x)\to0\), what additional ratio would you evaluate in order to find \(\lim_{x\to a}\frac{1-\cos(g(x))}{h(x)^2}\)?
- Why does the basic cosine defect limit require its input to approach zero?
