Answer:
The statement that correctly uses limits to determine the end behavior of f(x) is;
[tex]\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}= \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }[/tex] so the end behavior of the function is that as x → ±∞, f(x) → 0
Step-by-step explanation:
The given function is presented here as follows;
[tex]f(x) = \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}[/tex]
The limit of the function is presented as follows;
[tex]\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}[/tex]
Dividing the terms by x², we have;
[tex]\lim\limits_{x \to \pm \infty} \dfrac{\dfrac{7 \cdot x^2}{x^2} + \dfrac{x}{x^2} + \dfrac{1}{x^2} }{\dfrac{x^4}{x^2} +\dfrac{1}{x^2} }= \lim\limits_{x \to \pm \infty} \dfrac{7 + \dfrac{1}{x} + \dfrac{1}{x^2} }{x^2 +\dfrac{1}{x^2} }[/tex]
As 'x' tends to ±∞, we have;
[tex]\lim\limits_{x \to \pm \infty} \dfrac{7 + \dfrac{1}{x} + \dfrac{1}{x^2} }{x^2 +\dfrac{1}{x^2} } = \lim\limits_{x \to \pm \infty} \dfrac{7 + 0 + 0 }{x^2 +0 } = \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }[/tex]
However, we have that the end behavior of 7/x² as 'x' tends to ±∞ is 7/x² tends to 0;
Therefore, we have;
[tex]f(x) \rightarrow 0 \ as \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }[/tex]
The statement that correctly uses limits to determine the end behavior of f(x) is therefor given as follows;
[tex]\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}= \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }[/tex] so the end behavior of the function is that as x → ±∞, f(x) → 0.
