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mafiozo [28]
3 years ago
9

Need to show work for my answer working on limits

Mathematics
1 answer:
forsale [732]3 years ago
6 0

Answer:

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

General Formulas and Concepts:
<u>Calculus</u>

Limits

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

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify given.</em>

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

<u>Step 2: Evaluate</u>

  1. [Limit] Simplify:
    \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)
  2. [Limit] Apply Limit Rule [Variable Direct Substitution]:
    \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}

Recall that infinity in the denominator <em>converges</em> to 0. Therefore, we can focus on the first part:
\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}

∴ we have found the limit to <em>equal</em> negative infinity.

---

Learn more about limits: brainly.com/question/27517668

Learn more about calculus: brainly.com/question/27351658

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

Unit: Limits

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