Proof of Trigonometric Limit Properties

Sahabat Smaridasa - Mathematical Concepts

Previously, we posted material on "Solving Trigonometric Function Limits" using the properties of trigonometric limits. Now, we will study the Proof of Trigonometric Limit Properties, which is very useful for understanding trigonometric limits.

Proving these properties is important because if the properties are not valid, the results of trigonometric limits would be incorrect. Although the proofs are not trivial and involve trigonometric formulas, we have listed all the necessary formulas here. We hope these proofs are beneficial for learning trigonometric limits.

Required Theorems and Formulas

Squeeze Theorem (Theorem of Apit):

Let \(f, g, h\) be functions defined on an open interval \(I\) containing \(a\) (except possibly at \(a\) itself) such that \(f(x) \le g(x) \le h(x)\) for every \(x \in I, x \ne a\). If \(\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L\), then \(\lim_{x \to a} g(x) = L\).

Area Formulas:

• Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
• Area of sector (circle): \(\text{Area of sector } AOB = \frac{1}{2} \cdot \angle AOB \cdot r^2\)

Geometric Setup for the Proof

Figure: Circle with radius \(r\), central angle \(x\), points O (center), B, C, A, D. Here \(OB = OA = r\).

From the figure:

• In \(\triangle BOC\): \(\sin x = \frac{BC}{OB} = \frac{BC}{r} \Rightarrow BC = r \sin x\)
• In \(\triangle BOC\): \(\cos x = \frac{OC}{OB} = \frac{OC}{r} \Rightarrow OC = r \cos x\)
• In \(\triangle AOD\): \(\tan x = \frac{AD}{OA} = \frac{AD}{r} \Rightarrow AD = r \tan x\)
• Note: \(OB = OA = r\) (radius of the circle)

Proof of \(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1\)

📐 Proof 1: \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)

Step 1: Area inequalities
From the figure: \[ \text{Area of } \triangle BOC = \frac{1}{2} \cdot OC \cdot BC = \frac{1}{2} \cdot r \cos x \cdot r \sin x = \frac{1}{2}r^2 \cos x \sin x \] \[ \text{Area of sector } AOB = \frac{1}{2} \cdot \angle AOB \cdot r^2 = \frac{1}{2} r^2 x \] \[ \text{Area of } \triangle AOD = \frac{1}{2} \cdot OA \cdot AD = \frac{1}{2} \cdot r \cdot r \tan x = \frac{1}{2} r^2 \tan x \] Clearly: \(\text{Area}(\triangle BOC) \le \text{Area(sector)} \le \text{Area}(\triangle AOD)\). Thus: \[ \frac{1}{2}r^2 \cos x \sin x \le \frac{1}{2}r^2 x \le \frac{1}{2}r^2 \tan x \] Divide by \(\frac{1}{2}r^2 > 0\): \[ \cos x \sin x \le x \le \tan x \]

Step 2: Divide by \(\sin x > 0\) for small \(x>0\)
\[ \cos x \le \frac{x}{\sin x} \le \frac{1}{\cos x} \]

Step 3: Apply Squeeze Theorem
As \(x \to 0^+\), \(\cos x \to 1\) and \(\frac{1}{\cos x} \to 1\). Hence: \[ \lim_{x \to 0^+} \frac{x}{\sin x} = 1 \quad \Rightarrow \quad \lim_{x \to 0^+} \frac{\sin x}{x} = 1 \] The same holds for \(x \to 0^-\) (using symmetry). Therefore: \[ \boxed{\lim_{x \to 0} \frac{\sin x}{x} = 1} \]

Proof of \(\displaystyle \lim_{x \to 0} \frac{\tan x}{x} = 1\)

📐 Proof 2: \(\lim_{x \to 0} \frac{\tan x}{x} = 1\)

Using the inequality from Proof 1:
We had \(\cos x \sin x \le x \le \tan x\). Divide the inequality by \(\tan x\) (positive for small \(x>0\)): \[ \cos x \sin x \cdot \frac{1}{\tan x} \le \frac{x}{\tan x} \le 1 \] Since \(\frac{\sin x}{\tan x} = \cos x\), the left side becomes \(\cos^2 x\). Thus: \[ \cos^2 x \le \frac{x}{\tan x} \le 1 \] As \(x \to 0\), \(\cos^2 x \to 1\). By the Squeeze Theorem: \[ \lim_{x \to 0} \frac{x}{\tan x} = 1 \quad \Rightarrow \quad \boxed{\lim_{x \to 0} \frac{\tan x}{x} = 1} \]

Proof of \(\displaystyle \lim_{x \to 0} \frac{\sin ax}{ax} = 1\)

📐 Proof 3: \(\lim_{x \to 0} \frac{\sin ax}{ax} = 1\)

Let \(y = ax\). As \(x \to 0\), \(y \to 0\). Then: \[ \lim_{x \to 0} \frac{\sin ax}{ax} = \lim_{y \to 0} \frac{\sin y}{y} = 1 \] This proves the property.

Proof of \(\displaystyle \lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b}\)

📐 Proof 4: \(\lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b}\)

Rewrite the expression: \[ \lim_{x \to 0} \frac{\sin ax}{bx} = \lim_{x \to 0} \frac{\sin ax}{ax} \cdot \frac{a}{b} = 1 \cdot \frac{a}{b} = \frac{a}{b} \] Thus, \(\displaystyle \lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b}\).

Similarly, we can prove: \(\displaystyle \lim_{x \to 0} \frac{\tan ax}{bx} = \frac{a}{b}\), \(\displaystyle \lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b}\), \(\displaystyle \lim_{x \to 0} \frac{\tan ax}{\tan bx} = \frac{a}{b}\), \(\displaystyle \lim_{x \to 0} \frac{\sin ax}{\tan bx} = \frac{a}{b}\).

Note: The proofs above rely on the Squeeze Theorem and basic trigonometric identities. For the general case where \(f(k)=0\), similar reasoning yields: \[ \lim_{x \to k} \frac{\sin a f(x)}{b f(x)} = \frac{a}{b}, \quad \lim_{x \to k} \frac{\tan a f(x)}{b f(x)} = \frac{a}{b} \] and the other related forms.