Answers
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
[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][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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Learn more about limits: https://brainly.com/question/27517668
Learn more about calculus: https://brainly.com/question/27351658
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Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits
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=========================================================
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