lim_(x→0) (1 - cos x)/x² =
A1/2
B2
C1
D0
Answer & Solution
Correct answer: A. 1/2
Use 1 - cos x = 2 sin²(x/2). Then (1 - cos x)/x² = 2 × (sin(x/2)/x)² × 1 = (1/2) × (sin(x/2)/(x/2))². As x → 0, sin(x/2)/(x/2) → 1, so limit = 1/2.
Related questions
Two things this chain introduces beside continuity are differentiability and:Composites of continuous functions, by Theorem 2, are:A function that fails the limit test at a point is:To differentiate an implicit relation you use the:The function f(x) = |x| at x = 0 is continuous but not:For f(x) = 2x + 3 at x = 1, the limit and f(1) are both:An alternative to repeated product rule for many factors is:The chain rule can be extended to composites of: