Question

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Answer 1

Answer:

The statement that correctly uses limits to determine the end behavior of f(x) is;

$\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}= \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$ 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;

$f(x) = \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}$

The limit of the function is presented as follows;

$\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}$

Dividing the terms by x², we have;

$\lim\limits_{x \to \pm \infty} \dfrac{\dfrac{7 \cdot x^2}{x^2} + \dfrac{x}{x^2} + \dfrac{1}{x^2} }{\dfrac{x^4}{x^2} +\dfrac{1}{x^2} }= \lim\limits_{x \to \pm \infty} \dfrac{7 + \dfrac{1}{x} + \dfrac{1}{x^2} }{x^2 +\dfrac{1}{x^2} }$

As 'x' tends to ±∞, we have;

$\lim\limits_{x \to \pm \infty} \dfrac{7 + \dfrac{1}{x} + \dfrac{1}{x^2} }{x^2 +\dfrac{1}{x^2} } = \lim\limits_{x \to \pm \infty} \dfrac{7 + 0 + 0 }{x^2 +0 } = \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$

However, we have that the end behavior of 7/x² as 'x' tends to ±∞ is 7/x² tends to 0;

Therefore, we have;

$f(x) \rightarrow 0 \ as \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$

The statement that correctly uses limits to determine the end behavior of f(x) is therefor given as follows;

$\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}= \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$ so the end behavior of the function is that as x → ±∞, f(x) → 0.