Answer:
See explanation.
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Algebra I
- Functions
- Function Notation
Calculus
Limits
- Right-Side Limit: [tex]\displaystyle \lim_{x \to c^+} f(x)[/tex]
- Left-Side Limit: [tex]\displaystyle \lim_{x \to c^-} f(x)[/tex]
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle f(x) = \left \{ {{\sqrt{x + 1}, \ x < 3} \atop {5 - x, \ x \geq 3}} \right.[/tex]
Step 2: Find Right Limit
- Substitute in variables [Right-Side Limit]: [tex]\displaystyle \lim_{x \to 3^+} 5 - x[/tex]
- Evaluate limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to 3^+} 5 - x = 5 - 3[/tex]
- Subtract: [tex]\displaystyle \lim_{x \to 3^+} 5 - x = 2[/tex]
∴ the right-side limit equals 2.
Step 3: Find Left Limit
- Substitute in variables [Left-Side Limit]: [tex]\displaystyle \lim_{x \to 3^-} \sqrt{x + 1}[/tex]
- Evaluate limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to 3^-} \sqrt{x + 1} = \sqrt{3 + 1}[/tex]
- [√Radical] Add: [tex]\displaystyle \lim_{x \to 3^-} \sqrt{x + 1} = \sqrt{4}[/tex]
- [√Radical] Evaluate: [tex]\displaystyle \lim_{x \to 3^-} \sqrt{x + 1} = 2[/tex]
∴ the left-side limit equals 2.
Step 4: Find Limit
The right and left-side limits are equal.
∴ [tex]\displaystyle \lim_{x \to 3} f(x) = 2[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits
Book: College Calculus 10e
