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Trigonometric Identities in Limits

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Core Definition

When direct substitution into a trigonometric limit yields the indeterminate form 00\frac{0}{0}, we use fundamental Trigonometric Identities to rewrite the expression into an algebraically equivalent form that isolates and eliminates the zero-inducing factor.

Concept Explanation

Trigonometric functions like sin⁡(x)\sin(x), cos⁡(x)\cos(x), and tan⁡(x)\tan(x) often produce 00\frac{0}{0} when evaluated at key unit circle angles (such as x=0,π2,πx = 0, \frac{\pi}{2}, \pi).

Unlike algebraic polynomials, trigonometric expressions cannot always be factored using standard polynomial rules without first using an identity to establish a common trigonometric function (e.g. converting everything into sines and cosines).

The Three Primary Trigonometric Limit Strategies

Strategy 1: Pythagorean Identity Substitution

Replace squared terms using sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x or cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x to create a difference of squares that factors into binomials:

1−cos⁡2x=(1−cos⁡x)(1+cos⁡x)1 - \cos^2 x = (1 - \cos x)(1 + \cos x)
1−sin⁡2x=(1−sin⁡x)(1+sin⁡x)1 - \sin^2 x = (1 - \sin x)(1 + \sin x)

Strategy 2: Multiplying by Trigonometric Conjugates

When facing binomials like (1−cos⁡x)(1 - \cos x) or (1−sin⁡x)(1 - \sin x) in a numerator or denominator that equals zero, multiply numerator and denominator by the conjugate (1+cos⁡x)(1 + \cos x) or (1+sin⁡x)(1 + \sin x). This converts the single term into a squared term via difference of squares:

lim⁡x→01−cos⁡xdenominator⋅1+cos⁡x1+cos⁡x=lim⁡x→0sin⁡2xdenominator⋅(1+cos⁡x)\lim_{x \to 0} \frac{1 - \cos x}{\text{denominator}} \cdot \frac{1 + \cos x}{1 + \cos x} = \lim_{x \to 0} \frac{\sin^2 x}{\text{denominator} \cdot (1 + \cos x)}

Strategy 3: Converting to Sine and Cosine

Rewrite tangents, secants, cosecants, and cotangents in terms of basic sin⁡(x)\sin(x) and cos⁡(x)\cos(x) using quotient and reciprocal identities:

tan⁡x=sin⁡xcos⁡x,sec⁡x=1cos⁡x,csc⁡x=1sin⁡x,cot⁡x=cos⁡xsin⁡x\tan x = \frac{\sin x}{\cos x}, \quad \sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x}, \quad \cot x = \frac{\cos x}{\sin x}

Example Problems

Example Problem 1: Factoring via Pythagorean Substitution

Evaluate the limit:

lim⁡x→0sin⁡2x1−cos⁡x\lim_{x \to 0} \frac{\sin^2 x}{1 - \cos x}

Solution

Step 1: Check direct substitution

sin⁡2(0)1−cos⁡(0)=021−1=00(Indeterminate)\frac{\sin^2(0)}{1 - \cos(0)} = \frac{0^2}{1 - 1} = \frac{0}{0} \quad (\text{Indeterminate})

Step 2: Apply the Pythagorean identity sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x

lim⁡x→01−cos⁡2x1−cos⁡x\lim_{x \to 0} \frac{1 - \cos^2 x}{1 - \cos x}

Step 3: Factor as a difference of squares

1−cos⁡2x=(1−cos⁡x)(1+cos⁡x)1 - \cos^2 x = (1 - \cos x)(1 + \cos x)
lim⁡x→0(1−cos⁡x)(1+cos⁡x)1−cos⁡x\lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x)}{1 - \cos x}

Step 4: Cancel common zero factor and evaluate

lim⁡x→0(1+cos⁡x)=1+cos⁡(0)=1+1=2\lim_{x \to 0} (1 + \cos x) = 1 + \cos(0) = 1 + 1 = 2

Answer: 22

Example Problem 2: Limit at x=π2x = \frac{\pi}{2}

Evaluate the limit:

lim⁡x→π21−sin⁡2x2cos⁡x\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin^2 x}{2\cos x}

Solution

Step 1: Check direct substitution

1−sin⁡2(π/2)2cos⁡(π/2)=1−122(0)=00\frac{1 - \sin^2(\pi/2)}{2\cos(\pi/2)} = \frac{1 - 1^2}{2(0)} = \frac{0}{0}

Step 2: Apply identity 1−sin⁡2x=cos⁡2x1 - \sin^2 x = \cos^2 x

lim⁡x→π2cos⁡2x2cos⁡x\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{2\cos x}

Step 3: Cancel cos⁡x\cos x and evaluate

lim⁡x→π2cos⁡x2=cos⁡(π/2)2=02=0\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{2} = \frac{\cos(\pi/2)}{2} = \frac{0}{2} = 0

Answer: 00

Example Problem 3: Trigonometric Conjugate Method

Evaluate the limit:

lim⁡x→01−cos⁡xsin⁡2x\lim_{x \to 0} \frac{1 - \cos x}{\sin^2 x}

Solution

Step 1: Check direct substitution

1−cos⁡(0)sin⁡2(0)=1−10=00\frac{1 - \cos(0)}{\sin^2(0)} = \frac{1 - 1}{0} = \frac{0}{0}

Step 2: Replace sin⁡2x\sin^2 x with 1−cos⁡2x1 - \cos^2 x in the denominator

lim⁡x→01−cos⁡x1−cos⁡2x=lim⁡x→01−cos⁡x(1−cos⁡x)(1+cos⁡x)\lim_{x \to 0} \frac{1 - \cos x}{1 - \cos^2 x} = \lim_{x \to 0} \frac{1 - \cos x}{(1 - \cos x)(1 + \cos x)}

Step 3: Cancel (1−cos⁡x)(1 - \cos x) and evaluate

lim⁡x→011+cos⁡x=11+cos⁡(0)=11+1=12\lim_{x \to 0} \frac{1}{1 + \cos x} = \frac{1}{1 + \cos(0)} = \frac{1}{1 + 1} = \frac{1}{2}

Answer: 12\frac{1}{2}

Connections & References