Q.31 The value of lim_x→0[(cos 2x − cos 4x) / x²] is ____________.

Limit Evaluation

The limit
limx→0 (cos 2x − cos 4x)/x² evaluates to 6.

Detailed Solution

This is a 0/0 indeterminate form as x → 0, since cos(0) = 1. Apply the trigonometric identity:

cos A − cos B = −2 sin((A+B)/2) sin((A−B)/2)

Let A = 2x and B = 4x.

Substitute to get:

cos 2x − cos 4x = −2 sin((2x + 4x)/2) sin((2x − 4x)/2)

= −2 sin(3x) sin(−x)

= 2 sin(3x) sin x (since sin(−θ) = −sin θ)

The limit becomes:

limx→0 2 sin(3x) sin x / x²

= 2 limx→0 (sin(3x)/x ⋅ sin x/x)

= 2 limx→0 (3 sin(3x)/(3x) ⋅ sin x/x)

= 2 × 3 × 1 × 1 = 6

Check By Taylor Series

cos 2x = 1 − 2x² + O(x⁴)
cos 4x = 1 − 8x² + O(x⁴)

Numerator:

(1 − 2x²) − (1 − 8x²) = 6x² + O(x⁴)

Divide by x² → 6

Check by L’Hôpital Rule

First derivatives give 0/0. After second derivative:

Numerator → −4cos2x + 16cos4x → −4 + 16 = 12

Denominator → 2

Limit = 12 / 2 = 6

Common Wrong Answers

• 2 (from sin x / x only)
• 12 (before dividing)
• 0 (ignoring higher terms)

Quick Recap

2 sin(3x) sin x / x² → 6. Taylor confirms numerator ~6x².