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

Answer:

$\displaystyle \lim_{x \to 5} \frac{x^2 - 25}{x - 5} = 10$

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]:                                                                   $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \lim_{x \to 5} \frac{x^2 - 25}{x - 5}$

Step 2: Find Limit

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

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

Unit:  Limits