Calculus I · Limits and Continuity · lesson

Cosine Limits and Trigonometric Identities

Concept

Learning objectives

Use conjugates and identities to reduce limits involving $1-\cos x$ , $\sin^2x$ , or other trigonometric expressions to standard forms.

Cosine Limits and Trigonometric Identities

A first cosine limit

\boxed{\lim_{x\to0}\frac{1-\cos x}{x}=0}.

To show this, rationalize with $1+\cos x$ :

\begin{aligned} \frac{1-\cos x}{x} &\cdot\frac{1+\cos x}{1+\cos x}\\ &=\frac{1-\cos^2x}{x(1+\cos x)}\\ &=\frac{\sin^2x}{x(1+\cos x)}\\ &=\frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}. \end{aligned}

The first factor approaches $1$ , and the second approaches $0/2=0$ . Therefore the product approaches $0$ .

The second important cosine limit

\boxed{\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12}.

Again rationalize:

\begin{aligned} \frac{1-\cos x}{x^2} &=\frac{\sin^2x}{x^2(1+\cos x)}\\ &=\left(\frac{\sin x}{x}\right)^2\frac1{1+\cos x}. \end{aligned}

Taking limits gives

1^2\cdot\frac12=\frac12.

Guided walkthrough

Recognize the squared pattern

Evaluate

\lim_{x\to0}\frac{\sin^2x}{x^2}.

Show worked solution

Rewrite as a square:

\frac{\sin^2x}{x^2} =\left(\frac{\sin x}{x}\right)^2.

The inside approaches $1$ , so the square approaches

\boxed{1}.

Worked example

Identity plus a standard limit

Evaluate

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

Show worked solution

Create $(3x)^2$ in the denominator:

\frac{1-\cos(3x)}{x^2} =9\frac{1-\cos(3x)}{(3x)^2}.

As $x\to0$ , $3x\to0$ , so the standard cosine limit gives

9\cdot\frac12=\boxed{\frac92}.

Worked example

Exam-level: use an identity before the standard limit

Evaluate

\lim_{x\to0}\frac{\sin(2x)\sin(5x)}{x^2}.

Show worked solution

Separate the two standard factors:

\frac{\sin(2x)\sin(5x)}{x^2} =\left(\frac{\sin(2x)}{x}\right) \left(\frac{\sin(5x)}{x}\right).

Each factor can be rewritten:

\frac{\sin(2x)}{x}=2\frac{\sin(2x)}{2x}\to2,

\frac{\sin(5x)}{x}=5\frac{\sin(5x)}{5x}\to5.

Therefore the product approaches

\boxed{10}.

Exam note

Do not replace $\sin x$ by $x$ as an algebraic identity. The statement $\sin x\sim x$ near zero means their ratio approaches $1$ ; it does not mean the two expressions are equal for nonzero $x$ . Preserve the limit argument.

After the explanation

Use the section idea

Reading lens

Can the expression be rewritten around a known small-angle limit, with every scaling factor accounted for?

Mental model

The fundamental sine limit is a reusable local shape: other trigonometric limits work when you expose that shape through identities and scaling.

Decision

Look for a bounded oscillation times a shrinking factor, or rewrite the expression into sine-over-angle factors whose arguments match their denominators.

Common trap

The sine function is not equal to its angle; their ratio merely approaches one near zero, and that statement requires radian measure.

Check yourself

Mastery means you can mark every scaling factor before simplifying and can explain where the Squeeze Theorem enters the argument.

Source & rights

Original instruction with traceable references.

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