Find the limit. lim θ→0 cos(8θ) − 1 / sin(3θ)

Answer:[tex]\displaystyle \lim_{\theta \to 0} \frac{\cos (8\theta) - 1}{\sin (3\theta)} = 0[/tex]General Formulas and Concepts:Pre-CalculusUnit CircleCalculusLimitsLimit 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]Step-by-step explanation:We are given the limit:[tex]\displaystyle \lim_{\theta \to 0} \frac{\cos (8\theta) - 1}{\sin (3\theta)}[/tex]If we evaluate the limit how it is using limit rules, we get:[tex]\displaystyle \lim_{\theta \to 0} \frac{\cos (8\theta) - 1}{\sin (3\theta)} = \frac{0}{0}[/tex]We see we have an indeterminate form, so let's use L'Hopital's Rule:[tex]\displaystyle \lim_{\theta \to 0} \frac{\cos (8\theta) - 1}{\sin (3\theta)} = \lim_{\theta \to 0} \frac{-8\sin (8x)}{3\cos (3x)}[/tex]Evaluating the new limit using limit rules, we get:[tex]\displaystyle \lim_{\theta \to 0} \frac{-8\sin (8x)}{3\cos (3x)} = \frac{-8\sin(0)}{3\cos(0)}[/tex]Which simplifies to:[tex]\displaystyle \lim_{\theta \to 0} \frac{-8\sin (8x)}{3\cos (3x)} = 0[/tex]And we have our answer.Topic: AP Calculus AB/BC (Calculus I/I + II)Unit: Limits