Question

Please help with this calculus problem

Answer 1

`\displaystyle \lim_(x \to 0) (1 - \cos x)/(x) = 0`

General Formulas and Concepts:

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))`

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Explanation:

Step 1: Define

Identify

`\displaystyle \lim_(x \to 0) (1 - \cos x)/(x)`

Step 2: Find

1. Special Limit Rule [L'Hopital's Rule]:
   `\displaystyle \lim_(x \to 0) (1 - \cos x)/(x) = \lim_(x \to 0) ((1 - \cos x)')/((x)')`
2. Rewrite [Derivative Rule - Addition/Subtraction]:
   `\displaystyle \lim_(x \to 0) (1 - \cos x)/(x) = \lim_(x \to 0) ((1)' - (\cos x)')/((x)')`
3. Basic Power Rule:
   `\displaystyle \lim_(x \to 0) (1 - \cos x)/(x) = \lim_(x \to 0) (0 - (\cos x)')/(1)`
4. Simplify:
   `\displaystyle \lim_(x \to 0) (1 - \cos x)/(x) = \lim_(x \to 0) -(\cos x)'`
5. Trigonometric Differentiation:
   `\displaystyle \lim_(x \to 0) (1 - \cos x)/(x) = \lim_(x \to 0) \sin x`
6. Evaluate Limit [Limit Rule - Variable Direct Substitution]:
   `\displaystyle \lim_(x \to 0) (1 - \cos x)/(x) = 0`

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

Unit: Advanced Limit Techniques