Question

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

Answer 1

`\displaystyle \lim_(\theta \to 0) (\cos (8\theta) - 1)/(\sin (3\theta)) = 0`

General Formulas and Concepts:

Pre-Calculus

• Unit Circle

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
`\displaystyle \lim_(x \to c) x = c`

Special Limit Rule [L’Hopital’s Rule]:
`\displaystyle \lim_(x \to c) (f(x))/(g(x)) = \lim_(x \to c) (f'(x))/(g'(x))`

Explanation:

We are given the limit:

`\displaystyle \lim_(\theta \to 0) (\cos (8\theta) - 1)/(\sin (3\theta))`

If we evaluate the limit how it is using limit rules, we get:

`\displaystyle \lim_(\theta \to 0) (\cos (8\theta) - 1)/(\sin (3\theta)) = (0)/(0)`

We see we have an indeterminate form, so let's use L'Hopital's Rule:

`\displaystyle \lim_(\theta \to 0) (\cos (8\theta) - 1)/(\sin (3\theta)) = \lim_(\theta \to 0) (-8\sin (8x))/(3\cos (3x))`

Evaluating the new limit using limit rules, we get:

`\displaystyle \lim_(\theta \to 0) (-8\sin (8x))/(3\cos (3x)) = (-8\sin(0))/(3\cos(0))`

Which simplifies to:

`\displaystyle \lim_(\theta \to 0) (-8\sin (8x))/(3\cos (3x)) = 0`

And we have our answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits