Answer:
[tex]\displaystyle \lim_{x \to 2} \frac{x^8 - 256}{x - 2} = 1024[/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
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^8 - 256}{x - 2}[/tex]
Substituting in x = 2 using the limit rule, we have:
[tex]\displaystyle \lim_{x \to 2} \frac{x^8 - 256}{x - 2} = \frac{2^8 - 256}{2 - 2}[/tex]
Evaluating the result, we get an indeterminate form:
[tex]\displaystyle \lim_{x \to 2} \frac{x^8 - 256}{x - 2} = \frac{0}{0}[/tex]
Since we have an indeterminate form, let's apply L'Hopital's Rule. Differentiate both the numerator and denominator respectively:
[tex]\displaystyle \lim_{x \to 2} \frac{x^8 - 256}{x - 2} = \lim_{x \to 2} \frac{8x^7}{1}[/tex]
Simplify:
[tex]\displaystyle \lim_{x \to 2} \frac{x^8 - 256}{x - 2} = \lim_{x \to 2} 8x^7[/tex]
Evaluate the limit using the limit rule:
[tex]\displaystyle \lim_{x \to 2} 8x^7 = 8(2)^7[/tex]
Simplify:
[tex]\displaystyle \lim_{x \to 2} 8x^7 = 1024[/tex]
And we have our answer.
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits
