MathCastle
PrecalculusChallengeIntroduction to Limits

Introduction to Limits · Challenge problem

Given only the small-angle facts lim⁡x→0sin⁡xx=1\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1 and lim⁡x→01−cos⁡xx2=12\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^{2}}=\frac12, evaluate

lim⁡x→0tan⁡x−sin⁡xx3.\lim_{x\to0}\frac{\tan x-\sin x}{x^{3}}.

(Hint: factor sin⁡x\sin x out of the numerator first.)