Q:

please help with this calculus problem

Accepted Solution

A:
Answer:[tex]\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = 0[/tex]General Formulas and Concepts:CalculusLimitsLimit Rule [Variable Direct Substitution]:                                                     [tex]\displaystyle \lim_{x \to c} x = c[/tex]Special Limit Rule [L’Hopital’s Rule]:                                                                     [tex]\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}[/tex]DifferentiationDerivativesDerivative NotationDerivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]  Basic Power Rule:f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Step-by-step explanation:Step 1: DefineIdentify[tex]\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x}[/tex]Step 2: FindSpecial Limit Rule [L'Hopital's Rule]:                                                           [tex]\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{(1 - \cos x)'}{(x)'}[/tex]Rewrite [Derivative Rule - Addition/Subtraction]:                                       [tex]\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{(1)' - (\cos x)'}{(x)'}[/tex]Basic Power Rule:                                                                                         [tex]\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{0 - (\cos x)'}{1}[/tex]Simplify:                                                                                                         [tex]\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} -(\cos x)'[/tex]Trigonometric Differentiation:                                                                       [tex]\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \sin x[/tex]Evaluate Limit [Limit Rule - Variable Direct Substitution]:                           [tex]\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = 0[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)Unit: Advanced Limit Techniques