Question

Use properties of limits and algebraic methods to find the limit, if it exists. (if the limit is infinite, enter '∞' or '-∞', as appropriate. if the limit does not otherwise exist, enter dne.) lim x→5 x2 − 25 x − 5

Answer 1

`\displaystyle \lim_(x \to 5) (x^2 - 25)/(x - 5) = 10`

General Formulas and Concepts:

Algebra I

• Terms/Coefficients
• Factoring

Calculus

Limits

Limit Rule [Constant]:
`\displaystyle \lim_(x \to c) b = b`

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

Limit Property [Addition/Subtraction]:
`\displaystyle \lim_(x \to c) [f(x) \pm g(x)] = \lim_(x \to c) f(x) \pm \lim_(x \to c) g(x)`

Explanation:

Step 1: Define

Identify

`\displaystyle \lim_(x \to 5) (x^2 - 25)/(x - 5)`

Step 2: Find Limit

1. [Limit] Factor:
   `\displaystyle \lim_(x \to 5) (x^2 - 25)/(x - 5) = \lim_(x \to 5) ((x - 5)(x + 5))/(x - 5)`
2. [Limit] Simplify:
   `\displaystyle \lim_(x \to 5) (x^2 - 25)/(x - 5) = \lim_(x \to 5) x + 5`
3. [Limit] Evaluate [Limit Rule - Variable Direct Substitution]:
   `\displaystyle \lim_(x \to 5) (x^2 - 25)/(x - 5) = 5 + 5`
4. Simplify:
   `\displaystyle \lim_(x \to 5) (x^2 - 25)/(x - 5) = 10`

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

Unit: Limits