alleryUse L'Hôpital's rule: lim_(x→0) (sin x - x)/x³ =
A-1/6
B0
C1/6
D-1
Answer & Solution
Correct answer: A. -1/6
0/0 form.
**Differentiate top/bottom:** (cos x - 1)/(3x²) — still 0/0. Again: (-sin x)/(6x) — still 0/0. Again: (-cos x)/6
→ -1/6. (Or use Taylor: sin x = x - x³/6 + ..., so (sin x - x)/x³ → -1/6.)
Keep practising Unit 4-5: Applications of Derivatives & Curve Sketching
Browse all Unit 4-5: Applications of Derivatives & Curve Sketching questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
If the tangent to the curve $y = e^{2x}$ at point $(a, e^{2a})$ passes through origin, theIf $f(x) = x^x$ for $x > 0$, then $f$ has a minimum at $x = $:The function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on $[0, \infty)$ at:Find the **shortest distance** from origin to the curve $xy = 4$ (in the first quadrant):A rectangle is inscribed in a semicircle of radius $R$ with base on the diameter. The **maThe function $f(x) = x^3 - 3x + 1$ has critical points at:$f(x) = 2x^3 - 3x^2 - 12x + 5$. The **local maximum value** occurs at:Find the approximate value of $ qrt{25.5}$ using differentials (given $ qrt{25} = 5$):