Answer:
[tex]\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}[/tex]
General Formulas and Concepts:
Calculus
Limits
Limit Rule [Constant]: [tex]\displaystyle \lim_{x \to c} b = b[/tex]
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
Limit Property [Addition/Subtraction]: [tex]\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)[/tex]
L'Hopital's Rule
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Step-by-step explanation:
We are given the following limit:
[tex]\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}[/tex]
Let's substitute in x = -2 using the limit rule:
[tex]\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}[/tex]
Evaluating this, we arrive at an indeterminate form:
[tex]\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}[/tex]
Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:
[tex]\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}[/tex]
Substitute in x = -2 using the limit rule:
[tex]\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{3(-2)^2}{4(-2)^3}[/tex]
Evaluating this, we get:
[tex]\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}[/tex]
And we have our answer.
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits
