Infinite Limits of Trigonometric Functions

Sahabat Smaridasa - Mathematical Concepts

In this article, we will discuss Infinite Limits of Trigonometric Functions. This topic combines infinite limits and trigonometric function limits. To master this material, you should first be familiar with trigonometric function limits.

Infinity (\(\infty\)) as an angle in a trigonometric function is problematic because \(\sin \infty\), \(\cos \infty\), and \(\tan \infty\) are not well-defined. Instead, we use forms like \(\frac{1}{\infty} = 0\), so that \(\sin(0)=0\), \(\cos(0)=1\), etc. This approach aligns with trigonometric limit properties.

Note: This type of problem appeared in the SBMPTN 2017 mathematics exam (one question per code). Therefore, a thorough understanding of trigonometric limits is essential.

Key Formulas and Properties

Trigonometric Limit Properties:

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

Trigonometric Identities:

• \(1 - \cos px = 2 \sin^2 \frac{px}{2}\)
• \(\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}\)
• \(\sin^2 x + \cos^2 x = 1\)

Infinite Limit for Rational Functions: Compare the highest powers of numerator and denominator.

📘 Example 1 : Basic Substitution

Find the following limits:

a) \(\displaystyle \lim_{x \to \infty} x \tan \frac{1}{x}\)    b) \(\displaystyle \lim_{y \to \infty} \frac{1}{y} \cot \frac{1}{y}\)    c) \(\displaystyle \lim_{x \to \infty} \frac{\csc \frac{1}{x}}{x}\)

Solution (using substitution \(y = 1/x\)):

a) Let \(y = 1/x\). As \(x \to \infty\), \(y \to 0\). Then: \[ \lim_{x \to \infty} x \tan \frac{1}{x} = \lim_{y \to 0} \frac{\tan y}{y} = 1. \]

b) Let \(x = 1/y\). As \(y \to \infty\), \(x \to 0\). Then: \[ \lim_{y \to \infty} \frac{1}{y} \cot \frac{1}{y} = \lim_{x \to 0} x \cot x = \lim_{x \to 0} \frac{x}{\tan x} = 1. \]

c) Let \(y = 1/x\). Then: \[ \lim_{x \to \infty} \frac{\csc(1/x)}{x} = \lim_{y \to 0} \frac{y}{\sin y} = 1. \]

📘 Example 2 : Products of Trigonometric Functions

Find:

a) \(\displaystyle \lim_{x \to \infty} \tan \frac{5}{x} \cdot \csc \frac{2}{x}\)    b) \(\displaystyle \lim_{x \to \infty} \cot \frac{3}{x} \cdot \sin \frac{1}{x}\)    c) \(\displaystyle \lim_{x \to \infty} \frac{\cot \frac{1}{2x}}{\csc \frac{3}{x}}\)

Solution (substitute \(y = 1/x\)):

a) \(\displaystyle \lim_{x \to \infty} \tan \frac{5}{x} \cdot \csc \frac{2}{x} = \lim_{y \to 0} \frac{\tan 5y}{\sin 2y} = \frac{5}{2}\).

b) \(\displaystyle \lim_{x \to \infty} \cot \frac{3}{x} \cdot \sin \frac{1}{x} = \lim_{y \to 0} \frac{\sin y}{\tan 3y} = \frac{1}{3}\).

c) \(\displaystyle \lim_{x \to \infty} \frac{\cot \frac{1}{2x}}{\csc \frac{3}{x}} = \lim_{y \to 0} \frac{\sin 3y}{\tan(y/2)} = \frac{3}{1/2} = 6\).

📘 Example 3 : SBMPTN Style Problem

Find \(\displaystyle \lim_{y \to \infty} \sqrt{6y} \cos \frac{3}{\sqrt{y}} \sin \frac{5}{\sqrt{y}}\).

Solution: Let \(x = 1/\sqrt{y}\), so \(y \to \infty \Rightarrow x \to 0\). Then: \[ \sqrt{6y} = \sqrt{6} \cdot \frac{1}{x}, \quad \cos \frac{3}{\sqrt{y}} = \cos 3x, \quad \sin \frac{5}{\sqrt{y}} = \sin 5x. \] Thus: \[ \lim_{x \to 0} \sqrt{6} \cdot \frac{\sin 5x}{x} \cdot \cos 3x = \sqrt{6} \cdot 5 \cdot 1 = 5\sqrt{6}. \]

📘 Example 4 : Using \(1-\cos\) Identity

Find \(\displaystyle \lim_{x \to \infty} \frac{1 - \cos \frac{4}{x}}{\frac{1}{x} \cdot \tan \frac{3}{x}}\).

Solution: Let \(y = 1/x\). Then: \[ \lim_{y \to 0} \frac{1 - \cos 4y}{y \cdot \tan 3y} = \lim_{y \to 0} \frac{2 \sin^2 2y}{y \cdot \tan 3y} = \lim_{y \to 0} 2 \cdot \frac{\sin 2y}{y} \cdot \frac{\sin 2y}{\tan 3y} = 2 \cdot 2 \cdot \frac{2}{3} = \frac{8}{3}. \]

📘 Example 5 : More Complex Limit

Find \(\displaystyle \lim_{x \to \infty} \frac{(2x-3) \cot \frac{2}{x}}{5x-2}\).

Solution: Let \(y = 1/x\). Then: \[ \lim_{y \to 0} \frac{(2/y - 3) \cot 2y}{5/y - 2} = \lim_{y \to 0} \frac{(2 - 3y) \cdot (1/y) \cdot (1/\tan 2y)}{5 - 2y} = \lim_{y \to 0} \frac{(2-3y)}{5-2y} \cdot \frac{1}{y \tan 2y} = \frac{2}{5} \cdot \frac{1}{2} = \frac{1}{5}. \]

📘 Example 6 : Challenging SBMPTN Problem

Find \(\displaystyle \lim_{x \to \infty} \frac{\cos \frac{4}{x} + \cos \frac{2}{x} \cdot \sin \frac{3}{\sqrt{x}} - \cos \frac{4}{x} \cdot \sin \frac{3}{\sqrt{x}} - \cos \frac{2}{x}}{\sin^2 \frac{1}{x} - \cos \frac{2}{x} + 1}\).

Solution: Let \(y = 1/x\) and \(t = \sqrt{y}\). After simplification using identities: \[ \text{Numerator} = (\cos 4y - \cos 2y)(1 - \sin 3\sqrt{y}) = -2 \sin 3y \sin y \cdot (1 - \sin 3\sqrt{y}). \] Denominator: \(\sin^2 y + (1 - \cos 2y) = \sin^2 y + 2\sin^2 y = 3\sin^2 y.\) Thus: \[ \lim_{y \to 0} \frac{-2 \sin 3y \cdot \sin y \cdot (1 - \sin 3\sqrt{y})}{3\sin^2 y} = \lim_{y \to 0} \frac{-2 \sin 3y \cdot (1 - \sin 3\sqrt{y})}{3 \sin y} = \frac{-2}{3} \cdot \lim_{y \to 0} \frac{\sin 3y}{\sin y} \cdot \lim_{y \to 0} (1 - \sin 3\sqrt{y}) = \frac{-2}{3} \cdot 3 \cdot (1-0) = -2. \]

SBMPTN 2017 Sample Problems

1. (Kode 165) \(\displaystyle \lim_{y \to \infty} y \cdot \sin \frac{3}{y} \cdot \cos \frac{5}{y} = \ldots\)

2. (Kode 166) \(\displaystyle \lim_{x \to \infty} \frac{\sin \frac{3}{x}}{(1 - \cos \frac{2}{x}) \cdot x^2 \cdot \sin \frac{1}{x}} = \ldots\)

3. (Kode 167) \(\displaystyle \lim_{x \to \infty} x \left(1 - \cos \frac{1}{\sqrt{x}}\right) = \ldots\)

4. (Kode 168) \(\displaystyle \lim_{x \to \infty} 2x \tan \frac{1}{x} \cdot \sec \frac{2}{x} = \ldots\)