Word Count : 424

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We solve each part step by step. 1. Evaluate $$ \lim_{x\to 0} \frac{\tan x - \sin x}{x^3} $$ Use the Taylor series expansions around $x = 0$: $$ \sin x = x - \frac{x^3}{6} + \cdots, \quad \tan x = x + \frac{x^3}{3} + \cdots $$ Subtract: $$ \tan x - \sin x = \left(x + \frac{x^3}{3} + \cdots \right) - \left(x - \frac{x^3}{6} + \cdots \right) = \frac{x^3}{3} + \frac{x^3}{6} + \cdots = \frac{x^3}{2} + \cdots $$ Divide by $x^3$: $$ \frac{\tan x - \sin x}{x^3} = \frac{x^3/2 + \cdots}{x^3} \to \frac{1}{2} \quad \text{as } x \to 0 $$ So the limit is $$ \boxed{\frac{1}{2}} ______ _______ _______ __________ ______ ___ ____ _________ __________ _____ ___.
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