Answer :
Answer:
[tex]\displaystyle \lim_{x \to 5} \frac{x^2 - 25}{x + 5} = 0[/tex]
General Formulas and Concepts:
Algebra I
Terms/Coefficients
- Factoring
Calculus
Limits
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \lim_{x \to 5} \frac{x^2 - 25}{x + 5}[/tex]
Step 2: Evaluate
- Factor: [tex]\displaystyle \lim_{x \to 5} \frac{x^2 - 25}{x + 5} = \lim_{x \to 5} \frac{(x - 5)(x + 5)}{x + 5}[/tex]
- Simplify: [tex]\displaystyle \lim_{x \to 5} \frac{x^2 - 25}{x + 5} = \lim_{x \to 5} x - 5[/tex]
- Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to 5} \frac{x^2 - 25}{x + 5} = 5 - 5[/tex]
- Simplify: [tex]\displaystyle \lim_{x \to 5} \frac{x^2 - 25}{x + 5} = 0[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits