Answer :
To solve for the limits of the function as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] and [tex]\( -\infty \)[/tex], we need to examine the behavior of the function [tex]\( f(x) = \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} \)[/tex].
1. Limit as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} \][/tex]
When [tex]\( x \)[/tex] is very large (approaches infinity), the highest power of [tex]\( x \)[/tex] in the numerator ([tex]\( -2x^5 \)[/tex]) dominates, and similarly, the highest power in the denominator ([tex]\( -x^2 \)[/tex]) dominates. Therefore, the leading term in the numerator and the leading term in the denominator will primarily dictate the behavior of the fraction as [tex]\( x \)[/tex] becomes very large.
[tex]\[ \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} \approx \frac{-2x^5}{-x^2} = 2x^3 \][/tex]
As [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex],
[tex]\[ 2x^3 \to \infty \][/tex]
Hence,
[tex]\[ \lim_{x \to \infty} \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} = \infty \][/tex]
2. Limit as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex]:
[tex]\[ \lim_{x \to -\infty} \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} \][/tex]
Similarly, when [tex]\( x \)[/tex] is very large in magnitude but negative (approaches negative infinity), the highest power terms will still dominate:
[tex]\[ \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} \approx \frac{-2x^5}{-x^2} = 2x^3 \][/tex]
As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex],
[tex]\[ 2x^3 \to -\infty \][/tex]
Hence,
[tex]\[ \lim_{x \to -\infty} \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} = -\infty \][/tex]
So, the evaluated limits are:
[tex]\[ \lim_{x \to \infty} f(x) = \infty \][/tex]
[tex]\[ \lim_{x \to -\infty} f(x) = -\infty \][/tex]
Therefore, the limits are [tex]\( \infty \)[/tex] as [tex]\( x \to \infty \)[/tex] and [tex]\( -\infty \)[/tex] as [tex]\( x \to -\infty \)[/tex].
1. Limit as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} \][/tex]
When [tex]\( x \)[/tex] is very large (approaches infinity), the highest power of [tex]\( x \)[/tex] in the numerator ([tex]\( -2x^5 \)[/tex]) dominates, and similarly, the highest power in the denominator ([tex]\( -x^2 \)[/tex]) dominates. Therefore, the leading term in the numerator and the leading term in the denominator will primarily dictate the behavior of the fraction as [tex]\( x \)[/tex] becomes very large.
[tex]\[ \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} \approx \frac{-2x^5}{-x^2} = 2x^3 \][/tex]
As [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex],
[tex]\[ 2x^3 \to \infty \][/tex]
Hence,
[tex]\[ \lim_{x \to \infty} \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} = \infty \][/tex]
2. Limit as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex]:
[tex]\[ \lim_{x \to -\infty} \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} \][/tex]
Similarly, when [tex]\( x \)[/tex] is very large in magnitude but negative (approaches negative infinity), the highest power terms will still dominate:
[tex]\[ \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} \approx \frac{-2x^5}{-x^2} = 2x^3 \][/tex]
As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex],
[tex]\[ 2x^3 \to -\infty \][/tex]
Hence,
[tex]\[ \lim_{x \to -\infty} \frac{-2x^5 + 4x^3 + 2x}{-x^2 - 1} = -\infty \][/tex]
So, the evaluated limits are:
[tex]\[ \lim_{x \to \infty} f(x) = \infty \][/tex]
[tex]\[ \lim_{x \to -\infty} f(x) = -\infty \][/tex]
Therefore, the limits are [tex]\( \infty \)[/tex] as [tex]\( x \to \infty \)[/tex] and [tex]\( -\infty \)[/tex] as [tex]\( x \to -\infty \)[/tex].
