Solving Trigonometric Function Limits

Sahabat Smaridasa - Mathematical Concepts

After studying the material on "Solving Algebraic Function Limits", we will now continue with Trigonometric Function Limits. Here we will involve trigonometric functions, so we need to recall relevant trigonometry concepts such as trigonometric identities, sum and difference formulas, and double-angle formulas.

Solving trigonometric limits typically starts with direct substitution. If the result is an indeterminate form, we proceed using factorization, multiplying by the conjugate, using trigonometric limit properties, or using derivatives.

Determinate vs. Indeterminate Forms: As discussed in the algebraic limits article, if substitution yields \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), etc., further manipulation is required.

Properties of Trigonometric Limits

Basic Trigonometric Limit Properties:

• \(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1\)    and generally \(\displaystyle \lim_{x \to 0} \frac{\sin ax}{ax} = 1\)
• \(\displaystyle \lim_{x \to 0} \frac{\tan x}{x} = 1\)    and generally \(\displaystyle \lim_{x \to 0} \frac{\tan ax}{ax} = 1\)

Generalized Properties (for \(f(k)=0\)):

• \(\displaystyle \lim_{x \to k} \frac{\sin a f(x)}{b f(x)} = \frac{a}{b}\)
• \(\displaystyle \lim_{x \to k} \frac{\tan a f(x)}{b f(x)} = \frac{a}{b}\)
• \(\displaystyle \lim_{x \to k} \frac{\sin a f(x)}{\sin b f(x)} = \frac{a}{b}\)
• \(\displaystyle \lim_{x \to k} \frac{\tan a f(x)}{\tan b f(x)} = \frac{a}{b}\)
• \(\displaystyle \lim_{x \to k} \frac{\sin a f(x)}{\tan b f(x)} = \frac{a}{b}\)

These properties hold provided \(f(k)=0\).

📘 Example 1 : Basic Trigonometric Limits

Evaluate the following limits:

a) \(\displaystyle \lim_{x \to 0} \frac{\sin 3x}{5x}\)    b) \(\displaystyle \lim_{x \to 0} \frac{2x}{3 \sin 5x}\)    c) \(\displaystyle \lim_{x \to 0} \frac{7 \tan 2x}{4x}\)    d) \(\displaystyle \lim_{x \to 0} \frac{2x}{9 \tan 2x}\)

Solution (using generalized properties):

a) \(\displaystyle \lim_{x \to 0} \frac{\sin 3x}{5x} = \frac{3}{5}\).

b) \(\displaystyle \lim_{x \to 0} \frac{2x}{3 \sin 5x} = \frac{2}{3} \cdot \lim_{x \to 0} \frac{x}{\sin 5x} = \frac{2}{3} \cdot \frac{1}{5} = \frac{2}{15}\).

c) \(\displaystyle \lim_{x \to 0} \frac{7 \tan 2x}{4x} = 7 \cdot \frac{2}{4} = \frac{14}{4} = \frac{7}{2}\).

d) \(\displaystyle \lim_{x \to 0} \frac{2x}{9 \tan 2x} = \frac{2}{9} \cdot \frac{1}{2} = \frac{1}{9}\).

📘 Example 2 : Limits Involving sin and tan ratios

Evaluate:

a) \(\displaystyle \lim_{x \to 0} \frac{\sin 2x}{\sin 3x}\)    b) \(\displaystyle \lim_{x \to 0} \frac{\tan 6x}{\tan 2x}\)    c) \(\displaystyle \lim_{x \to 0} \frac{\sin 3x}{\tan 2x}\)    d) \(\displaystyle \lim_{x \to 0} \frac{\tan 4x}{\sin 8x}\)    e) \(\displaystyle \lim_{x \to 0} 3x \cot 7x\)

Solution:

a) \(\displaystyle \lim_{x \to 0} \frac{\sin 2x}{\sin 3x} = \frac{2}{3}\).

b) \(\displaystyle \lim_{x \to 0} \frac{\tan 6x}{\tan 2x} = \frac{6}{2} = 3\).

c) \(\displaystyle \lim_{x \to 0} \frac{\sin 3x}{\tan 2x} = \frac{3}{2}\).

d) \(\displaystyle \lim_{x \to 0} \frac{\tan 4x}{\sin 8x} = \frac{4}{8} = \frac{1}{2}\).

e) \(\displaystyle \lim_{x \to 0} 3x \cot 7x = \lim_{x \to 0} \frac{3x}{\tan 7x} = \frac{3}{7}\).

📘 Example 3 : Limits with cos (using identities)

Evaluate:

a) \(\displaystyle \lim_{x \to 0} \frac{2 - 2 \cos 2x}{3x^2}\)    b) \(\displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\cos 2x}{x - \frac{\pi}{4}}\)    c) \(\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x \sin 2x}\)    d) \(\displaystyle \lim_{x \to 0} \frac{3x + \sin 2x}{5x}\)

Solution:

a) Using \(\cos 2x = 1 - 2\sin^2 x\): \[ \lim_{x \to 0} \frac{2 - 2(1 - 2\sin^2 x)}{3x^2} = \lim_{x \to 0} \frac{4\sin^2 x}{3x^2} = \frac{4}{3}. \]

b) Let \(p = x - \frac{\pi}{4}\). Then \(x = p + \frac{\pi}{4}\), and \(\cos 2x = \cos(2p + \frac{\pi}{2}) = -\sin 2p\). \[ \lim_{p \to 0} \frac{-\sin 2p}{p} = -2. \]

c) Using \(1 - \cos x = 2\sin^2\frac{x}{2}\): \[ \lim_{x \to 0} \frac{2\sin^2\frac{x}{2}}{x \sin 2x} = \lim_{x \to 0} \frac{2 \cdot \frac{x^2}{4}}{x \cdot 2x} = \frac{1}{4} \text{ (after simplication)}. \]

d) \(\displaystyle \lim_{x \to 0} \left( \frac{3x}{5x} + \frac{\sin 2x}{5x} \right) = \frac{3}{5} + \frac{2}{5} = 1\).

📘 Example 4 : Limits at Infinity and special forms

Evaluate:

a) \(\displaystyle \lim_{x \to \infty} x \sin \frac{1}{x}\)    b) \(\displaystyle \lim_{x \to 0} \frac{\tan x - \sin x}{4x^3}\)

Solution:

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

b) Using \(\tan x = \frac{\sin x}{\cos x}\) and \(1 - \cos x = 2\sin^2\frac{x}{2}\): \[ \lim_{x \to 0} \frac{\sin x(1 - \cos x)}{4x^3 \cos x} = \lim_{x \to 0} \frac{\sin x \cdot 2\sin^2\frac{x}{2}}{4x^3 \cos x} = \frac{1}{8}. \]

📘 Example 5 : Algebraic & Trigonometric Combined

Evaluate:

a) \(\displaystyle \lim_{x \to 1} \frac{\sin(x-1)}{x^2 - 1}\)    b) \(\displaystyle \lim_{x \to 2} \frac{x^2 + x - 6}{\tan(x-2)}\)    c) \(\displaystyle \lim_{x \to 0} \frac{\sin x}{\sqrt{4+x} - 2}\)

Solution:

a) Factor denominator: \(x^2-1 = (x-1)(x+1)\). \[ \lim_{x \to 1} \frac{\sin(x-1)}{(x-1)(x+1)} = 1 \cdot \frac{1}{1+1} = \frac{1}{2}. \]

b) Factor numerator: \(x^2+x-6 = (x-2)(x+3)\). \[ \lim_{x \to 2} \frac{(x-2)(x+3)}{\tan(x-2)} = 1 \cdot (2+3) = 5. \]

c) Multiply numerator and denominator by the conjugate \(\sqrt{4+x}+2\): \[ \lim_{x \to 0} \frac{\sin x}{x} \cdot (\sqrt{4+x}+2) = 1 \cdot (2+2) = 4. \]

📘 Example 6 : Limit Definition of Derivative for \(f(x) = \cos x\)

Find \(\displaystyle \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\) for \(f(x) = \cos x\).

Solution:

Using \(\cos(x+h) = \cos x \cos h - \sin x \sin h\): \[ \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h} = \lim_{h \to 0} \frac{\cos x(\cos h - 1) - \sin x \sin h}{h}. \] Using \(\cos h - 1 = -2\sin^2\frac{h}{2}\) and known limits: \[ = \cos x \cdot \lim_{h \to 0} \frac{-2\sin^2\frac{h}{2}}{h} - \sin x \cdot 1 = \cos x \cdot 0 - \sin x = -\sin x. \] Thus, the derivative of \(\cos x\) is \(-\sin x\).