Special Trigonometric Limits

Navigation: Wiki Home > Skills > Special Trigonometric Limits

The Two Limits That Make Calculus Work for Trig

Why does the derivative of $\sin x$ equal $\cos x$? The answer depends on two remarkable limits that emerge from the geometry of the unit circle. Without them, we couldn't differentiate any trigonometric function.

These are not just facts to memorize. They are the foundation of all trig derivatives. Master them here, and the formulas in the next skill will make perfect sense.

Prerequisite Map

Quick Reference

Property	Value
Concept	Limits
Chapter	2.4
Difficulty	Intermediate
Time	~20 minutes

The Two Fundamental Limits

The Sine Limit

$$\boxed{\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1}$$

Important: This only works when $\theta$ is measured in radians.

The Cosine Limit

$$\boxed{\lim_{\theta \to 0} \frac{\cos \theta - 1}{\theta} = 0}$$

Visualization: Why sin(θ)/θ → 1

                    B
                   /|
                  / |
                 /  |  arc length = θ
                /   |
               /θ   | sin θ
              /     |
             O──────A
               cos θ

As θ → 0, the arc AB and the height |BC| = sin θ
become nearly equal. Since arc = θ (in radians):

        sin θ ≈ θ  for small θ

For small angles, the sine of an angle (the vertical drop) is almost equal to the arc length (the angle in radians). This is why $\frac{\sin\theta}{\theta} \to 1$.

Proof of the Sine Limit (Squeeze Theorem)

For $0 < \theta < \pi/2$, geometry of the unit circle gives:

$$\sin \theta < \theta < \tan \theta$$

Divide by $\sin \theta$ (positive for these $\theta$):

$$1 < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta}$$

Take reciprocals (flipping the inequalities):

$$\cos \theta < \frac{\sin \theta}{\theta} < 1$$

As $\theta \to 0^+$:

• $\cos \theta \to 1$
• $1 \to 1$

By the Squeeze Theorem: $\lim_{\theta \to 0^+} \frac{\sin \theta}{\theta} = 1$

Since $\frac{\sin \theta}{\theta}$ is an even function (both numerator and denominator are odd), the left-hand limit equals the right-hand limit.

Therefore: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

Proof of the Cosine Limit

Multiply numerator and denominator by $\cos \theta + 1$:

$$\frac{\cos \theta - 1}{\theta} = \frac{(\cos \theta - 1)(\cos \theta + 1)}{\theta(\cos \theta + 1)} = \frac{\cos^2 \theta - 1}{\theta(\cos \theta + 1)}$$

Using $\cos^2\theta - 1 = -\sin^2\theta$:

$$= \frac{-\sin^2 \theta}{\theta(\cos \theta + 1)} = -\frac{\sin \theta}{\theta} \cdot \frac{\sin \theta}{\cos \theta + 1}$$

As $\theta \to 0$:

• $\frac{\sin \theta}{\theta} \to 1$
• $\frac{\sin \theta}{\cos \theta + 1} \to \frac{0}{1 + 1} = 0$

Therefore: $\lim_{\theta \to 0} \frac{\cos \theta - 1}{\theta} = -1 \cdot 0 = 0$

Using These Limits

The Key Technique: Make It Match

When you see $\frac{\sin(\text{stuff})}{\text{stuff}}$, you want the "stuff" to match exactly.

Example: Evaluate $\lim_{x \to 0} \frac{\sin 5x}{3x}$

Strategy: We need $\frac{\sin 5x}{5x}$, so multiply and divide by the right constant:

$$\frac{\sin 5x}{3x} = \frac{5}{3} \cdot \frac{\sin 5x}{5x}$$

As $x \to 0$, we have $5x \to 0$, so $\frac{\sin 5x}{5x} \to 1$.

$$\lim_{x \to 0} \frac{\sin 5x}{3x} = \frac{5}{3} \cdot 1 = \frac{5}{3}$$

Common Patterns

Limit Form	Strategy	Result
$\frac{\sin ax}{bx}$	Write as $\frac{a}{b} \cdot \frac{\sin ax}{ax}$	$\frac{a}{b}$
$\frac{\tan \theta}{\theta}$	Write as $\frac{\sin\theta}{\theta} \cdot \frac{1}{\cos\theta}$	$1$
$x \cot x$	Write as $\frac{x\cos x}{\sin x} = \frac{\cos x}{\sin x / x}$	$1$
$\frac{1 - \cos\theta}{\theta}$	Same as $\frac{\cos\theta - 1}{\theta}$ but negative	$0$

Practice Problems

Conceptual Check (CCI-Style)

Mastery Checklist

• [ ] I can state both fundamental trig limits from memory
• [ ] I can explain why $\frac{\sin\theta}{\theta} \to 1$ using the Squeeze Theorem
• [ ] I can evaluate limits of the form $\frac{\sin ax}{bx}$ by adjusting coefficients
• [ ] I can handle tangent limits by writing $\tan = \sin/\cos$
• [ ] I understand why these limits only work in radians

Mental Model

The small-angle approximations:

For tiny angles (in radians):

• $\sin\theta \approx \theta$
• $\cos\theta \approx 1$

So $\frac{\sin\theta}{\theta} \approx 1$ and $\frac{\cos\theta - 1}{\theta} \approx \frac{0}{\theta} \approx 0$.

These approximations become exact in the limit as $\theta \to 0$.

---

Connections

Looking back:

• Squeeze Theorem provides the proof technique
• Limit Laws allow algebraic manipulation

Looking ahead:

• Trig Derivative Formulas uses these limits to prove $\frac{d}{dx}\sin x = \cos x$
• These limits appear in Taylor series for sine and cosine

Real-world connections:

• Small-angle approximations in physics (pendulums, optics)
• Engineering calculations where $\sin\theta \approx \theta$ simplifies analysis

---

Previous	Up	Next
Squeeze Theorem	Skills Index	Trig Derivative Formulas

---

Last updated: 2026-01-22