Respuesta :

Space

Answer:

[tex]\displaystyle \lim_{x \to \infty} \frac{-4x^8 + x^3 - x + 7}{x^4} = - \infty[/tex]

General Formulas and Concepts:
Calculus

Limits

Limit Rule [Variable Direct Substitution]:
[tex]\displaystyle \lim_{x \to c} x = c[/tex]

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \lim_{x \to \infty} \frac{-4x^8 + x^3 - x + 7}{x^4}[/tex]

Step 2: Evaluate

  1. [Limit] Simplify:
    [tex]\displaystyle \lim_{x \to \infty} \frac{-4x^8 + x^3 - x + 7}{x^4} = \lim_{x \to \infty} \bigg( -4x^4 + \frac{1}{x} - \frac{1}{x^3} + \frac{7}{x^4} \bigg)[/tex]
  2. [Limit] Apply Limit Rule [Variable Direct Substitution]:
    [tex]\displaystyle \lim_{x \to \infty} \frac{-4x^8 + x^3 - x + 7}{x^4} = -4(\infty)^4 + \frac{1}{\infty} - \frac{1}{(\infty)^3} + \frac{7}{(\infty)^4}[/tex]

Recall that infinity in the denominator converges to 0. Therefore, we can focus on the first part:
[tex]\displaystyle \begin{aligned}\lim_{x \to \infty} \frac{-4x^8 + x^3 - x + 7}{x^4} & = -4(\infty)^4 + \frac{1}{\infty} - \frac{1}{(\infty)^3} + \frac{7}{(\infty)^4} \\& = -4(\infty)^4 \\& = -4(\infty) \\& = \boxed{- \infty} \\\end{aligned}[/tex]

∴ we have found the limit to equal negative infinity.

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

Unit: Limits

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