Answer:
[tex]\displaystyle \lim_{\theta \to 0} \frac{\cos (8\theta) - 1}{\sin (3\theta)} = 0[/tex]
General Formulas and Concepts:
Pre-Calculus
Calculus
Limits
Limit 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
