Limit of Trigonometric Function

When Substitution Resolves a Trigonometric Limit

Trigonometric limits ask the same local question as algebraic limits: what value does an expression approach near a chosen input? Sine and cosine are continuous everywhere, while tangent is continuous wherever cosine is nonzero, so direct substitution still works whenever the resulting expression is defined.

The special work begins when substitution produces an indeterminate form such as $\frac{0}{0}$. Near zero, use the fundamental sine limit and familiar trigonometric identities to rewrite the expression into a form whose limit can be evaluated.

Sine Limit at Zero

When an angle measured in radians approaches zero, the ratio of its sine to the angle approaches one. Measuring the angle in radians is what makes the value exactly one, and that single fact decides the value of every ratio built from sine and cosine.

Basic Sine Limit

The limit is:

The expression is undefined at $x=0$, so direct substitution does not establish the limit. A geometric proof uses the unit circle and the squeeze theorem.

In this theorem, measure $x$ in radians. Using degrees gives a different result.

Consequences of the Basic Limit

Use this limit to derive the following results:

Because $\tan x=\frac{\sin x}{\cos x}$ and $\cos x\to1$:

Use the identity $1-\cos x=2\sin^2\frac x2$:

Use the identity $1-\cos x=2\sin^2\frac x2$:

The identity gives both cosine limits directly, without using derivative rules that are developed later:

Properties of Trigonometric Limits

The fundamental limit determines several related ratios, because a constant factor inside the angle moves out of the limit and the reciprocal of the ratio behaves the same way.

Trigonometric Ratios

For real constants $a$ and $b \neq 0$, all following limits are taken as $x \to 0$:

Trigonometric Combinations

From the trigonometric ratio properties, we can derive several trigonometric combination properties:

Techniques for Solving Trigonometric Limits

Some trigonometric limits resolve by substitution, and others need an identity or a known limit. The expression itself tells you which case applies, because a sum or a difference inside the trigonometric function is the sign that an identity is needed.

Substitution and Manipulation Techniques

When facing complex trigonometric limits, we often need to manipulate expressions to use the fundamental theorem.

Calculate: $\lim_{x \to 0} \frac{\sin 2x}{3x}$

We manipulate to obtain the standard form:

Trigonometric Identity Techniques

Often we need to use trigonometric identities to simplify expressions. Always ensure the function is defined at the point being approached.

Calculate: $\lim_{x \to \frac{\pi}{4}} \frac{\sin x + \cos x}{\sin x}$

Since $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \neq 0$, the denominator is not zero at $x = \frac{\pi}{4}$.

Direct substitution:

Techniques for Special Forms

For limits involving the form $\frac{0}{0}$, we need special techniques.

Calculate: $\lim_{x \to \frac{\pi}{2}} \frac{\sin^2(x - \frac{\pi}{2})}{x(x - \frac{\pi}{2})}$

Let $u = x - \frac{\pi}{2}$, then when $x \to \frac{\pi}{2}$, we have $u \to 0$ and $x = u + \frac{\pi}{2}$.

Trigonometric Limits with Angle Identities

A sum or difference formula rewrites the expression into limits that are already known. Expanding the sum or difference produces separate sine and cosine terms, and each of those returns to the fundamental limit.

Using Sum and Difference Formulas

When dealing with trigonometric functions involving sum or difference of angles, we can use trigonometric identities.

Calculate: $\lim_{t \to 0} \frac{\cot 5t}{\cot 10t}$

Using the definition of cotangent:

Exercises

Each problem asks for a trigonometric limit. Decide whether substitution is enough or an identity is needed before calculating. Write the limit you use at every step, and check the result against the fundamental limit whenever the angle is scaled.

1. Calculate $\lim_{x \to 0} \frac{\sin 3x}{4x}$

2. Calculate $\lim_{x \to 0} \frac{\tan 2x}{\sin 5x}$

3. Calculate $\lim_{x \to 0} \frac{1 - \cos 3x}{x^2}$

4. Calculate $\lim_{x \to \frac{\pi}{6}} \frac{\sin x - \frac{1}{2}}{x - \frac{\pi}{6}}$

5. Calculate $\lim_{x \to 0} \frac{\sin x \cos x}{x}$

Worked Solutions

1. Solution:

   Use algebraic manipulation to obtain the standard form:

   $$\begin{aligned} \lim_{x \to 0} \frac{\sin 3x}{4x} &= \lim_{x \to 0} \frac{3 \sin 3x}{3 \cdot 4x} \\ &= \lim_{x \to 0} \frac{3}{4} \cdot \frac{\sin 3x}{3x} \end{aligned}$$

   $$= \frac{3}{4} \lim_{x \to 0} \frac{\sin 3x}{3x} = \frac{3}{4} \cdot 1 = \frac{3}{4}$$

2. Solution:

   Use trigonometric ratio properties:

   $$\begin{aligned} &\lim_{x \to 0} \frac{\tan 2x}{\sin 5x} \\ &= \lim_{x \to 0} \frac{\tan 2x}{2x} \cdot \frac{5x}{\sin 5x} \cdot \frac{2x}{5x} \end{aligned}$$

   $$= \lim_{x \to 0} \frac{\tan 2x}{2x} \cdot \lim_{x \to 0} \frac{5x}{\sin 5x} \cdot \frac{2}{5}$$

   $$= 1 \cdot 1 \cdot \frac{2}{5} = \frac{2}{5}$$

3. Solution:

   Use the identity $1 - \cos ax = 2\sin^2\frac{ax}{2}$:

   $$\lim_{x \to 0} \frac{1 - \cos 3x}{x^2} = \lim_{x \to 0} \frac{2\sin^2\frac{3x}{2}}{x^2}$$

   $$= \lim_{x \to 0} 2 \cdot \left(\frac{\sin\frac{3x}{2}}{\frac{3x}{2}}\right)^2 \cdot \left(\frac{3}{2}\right)^2$$

   $$= 2 \cdot 1^2 \cdot \frac{9}{4} = \frac{9}{2}$$

4. Solution:

   Let $h = x - \frac{\pi}{6}$, then $x = h + \frac{\pi}{6}$ and when $x \to \frac{\pi}{6}$, we have $h \to 0$:

   $$\lim_{h \to 0} \frac{\sin(h + \frac{\pi}{6}) - \frac{1}{2}}{h}$$

   $$= \lim_{h \to 0} \frac{\sin h \cos\frac{\pi}{6} + \cos h \sin\frac{\pi}{6} - \frac{1}{2}}{h}$$

   $$= \lim_{h \to 0} \frac{\sin h \cdot \frac{\sqrt{3}}{2} + \cos h \cdot \frac{1}{2} - \frac{1}{2}}{h}$$

   $$= \lim_{h \to 0} \frac{\frac{\sqrt{3}}{2}\sin h + \frac{1}{2}(\cos h - 1)}{h}$$

   $$= \frac{\sqrt{3}}{2} \lim_{h \to 0} \frac{\sin h}{h} + \frac{1}{2} \lim_{h \to 0} \frac{\cos h - 1}{h}$$

   $$= \frac{\sqrt{3}}{2} \cdot 1 + \frac{1}{2} \cdot 0 = \frac{\sqrt{3}}{2}$$

5. Solution:

   Use the identity $\sin x \cos x = \frac{1}{2}\sin 2x$:

   $$\lim_{x \to 0} \frac{\sin x \cos x}{x} = \lim_{x \to 0} \frac{\frac{1}{2}\sin 2x}{x}$$

   $$= \frac{1}{2} \lim_{x \to 0} \frac{\sin 2x}{x} = \frac{1}{2} \lim_{x \to 0} \frac{2 \sin 2x}{2x}$$

   $$= \frac{1}{2} \cdot 2 \lim_{x \to 0} \frac{\sin 2x}{2x} = 1 \cdot 1 = 1$$