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:
Calculus

Limits

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

Step-by-step explanation:

Step 1: Define

Identify given.

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

Step 2: Evaluate

[Limit] Simplify:
$\displaystyle \lim_{x \to \infty} \frac{-4x^8 + x^3 - x + 7}{x^4} = \lim_{x \to \infty} \bigg( -4x^4 + \frac{1}{x} - \frac{1}{x^3} + \frac{7}{x^4} \bigg)$
[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 converges to 0. Therefore, we can focus on the first part:
$\displaystyle \begin{aligned}\lim_{x \to \infty} \frac{-4x^8 + x^3 - x + 7}{x^4} & = -4(\infty)^4 + \frac{1}{\infty} - \frac{1}{(\infty)^3} + \frac{7}{(\infty)^4} \\& = -4(\infty)^4 \\& = -4(\infty) \\& = \boxed{- \infty} \\\end{aligned}$