To determine the end behavior of the function [tex]\( f(x) = \frac{x-5}{5x^3 - 2x^2 + 3} \)[/tex], we need to evaluate the limits as [tex]\( x \)[/tex] approaches both [tex]\( -\infty \)[/tex] and [tex]\( \infty \)[/tex].
Let's start by analyzing the function as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex]:
1. As [tex]\( x \to -\infty \)[/tex], the dominant term in the numerator is [tex]\( x \)[/tex] because it is of the highest degree (power 1).
2. The dominant term in the denominator is [tex]\( 5x^3 \)[/tex] because it is of the highest degree (power 3).
Since we are dealing with limits at infinity, we focus on the highest-degree terms:
[tex]\[ f(x) \approx \frac{x}{5x^3} = \frac{1}{5x^2} \][/tex]
As [tex]\( x \to -\infty \)[/tex], [tex]\( x^2 \)[/tex] grows very large, and thus [tex]\( \frac{1}{5x^2} \to 0 \)[/tex].
Therefore:
[tex]\[ \lim_{x \to -\infty} f(x) = 0 \][/tex]
Next, let's analyze the function as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex]:
1. Similar to the previous case, the dominant term in the numerator is [tex]\( x \)[/tex].
2. The dominant term in the denominator is [tex]\( 5x^3 \)[/tex].
Again, focusing on the highest-degree terms:
[tex]\[ f(x) \approx \frac{x}{5x^3} = \frac{1}{5x^2} \][/tex]
As [tex]\( x \to \infty \)[/tex], [tex]\( x^2 \)[/tex] grows very large, and thus [tex]\( \frac{1}{5x^2} \to 0 \)[/tex].
Therefore:
[tex]\[ \lim_{x \to \infty} f(x) = 0 \][/tex]
Summarizing the end behavior:
- [tex]\( f(x) \rightarrow 0 \)[/tex] as [tex]\( x \rightarrow -\infty \)[/tex]
- [tex]\( f(x) \rightarrow 0 \)[/tex] as [tex]\( x \rightarrow \infty \)[/tex]
Thus, the end behavior of [tex]\( f(x) \)[/tex] is that it approaches 0 as [tex]\( x \)[/tex] tends to both [tex]\( -\infty \)[/tex] and [tex]\( \infty \)[/tex].
