Question

Limx→0 (1 - cos3x)/(x^2)

Note: We are NOT allowed to use L'Hopital's Rule. Only algebraic manipulation, trig formulas, and limit laws.

Answer 1

9/2

Explanation:

`\lim_(x \to 0) (1-cos(3x))/(x^2)`

Use the half-angle identity 2 sin²(u/2) = 1 − cos u.

`\lim_(x \to 0) (2sin^2(3x/2))/(x^2)`

Multiply top and bottom by 9/4.

`\lim_(x \to 0) [(9/4)/(9/4) * (2sin^2(3x/2))/(x^2)]\\\lim_(x \to 0) (9/2\ sin^2(3x/2))/(9/4\ x^2)\\9/2\ \lim_(x \to 0) (sin^2(3x/2))/((3x/2)^2)\\9/2\ \lim_(x \to 0) ((sin(3x/2))/(3x/2))^2\\9/2\ (\lim_(x \to 0) (sin(3x/2))/(3x/2))^2`

Finally, use the identity lim(x→0) [(sin u) / u] = 1.

`9/2\ (1)^2\\9/2`