consider f(x) = (7x²+x + 1)/(x⁴+1). which statement correctly uses limits to determine the end - behavior of…

Explanation:

Step1: Divide numerator and denominator by highest - power of x

When finding the limit as $x\to\pm\infty$ of the rational function $f(x)=\frac{7x^{2}+x + 1}{x^{4}+1}$, we divide both the numerator and the denominator by $x^{4}$ (the highest - power of $x$ in the denominator). $\lim_{x\to\pm\infty}\frac{7x^{2}+x + 1}{x^{4}+1}=\lim_{x\to\pm\infty}\frac{\frac{7x^{2}}{x^{4}}+\frac{x}{x^{4}}+\frac{1}{x^{4}}}{\frac{x^{4}}{x^{4}}+\frac{1}{x^{4}}}=\lim_{x\to\pm\infty}\frac{\frac{7}{x^{2}}+\frac{1}{x^{3}}+\frac{1}{x^{4}}}{1 + \frac{1}{x^{4}}}$ As $x\to\pm\infty$, $\frac{1}{x^{n}}\to0$ for $n>0$. So, $\lim_{x\to\pm\infty}\frac{\frac{7}{x^{2}}+\frac{1}{x^{3}}+\frac{1}{x^{4}}}{1+\frac{1}{x^{4}}}=\frac{0 + 0+0}{1 + 0}=0$. Another way is to consider the degrees of the numerator and denominator. The degree of the numerator $n = 2$ and the degree of the denominator $m=4$. When $n<m$, $\lim_{x\to\pm\infty}\frac{a_{n}x^{n}+a_{n - 1}x^{n - 1}+\cdots+a_{0}}{b_{m}x^{m}+b_{m - 1}x^{m - 1}+\cdots+b_{0}} = 0$. We can also simplify as follows: $\lim_{x\to\pm\infty}\frac{7x^{2}+x + 1}{x^{4}+1}=\lim_{x\to\pm\infty}\frac{7x^{2}(1+\frac{1}{7x}+\frac{1}{7x^{2}})}{x^{4}(1+\frac{1}{x^{4}})}=\lim_{x\to\pm\infty}\frac{7}{x^{2}}\cdot\frac{1+\frac{1}{7x}+\frac{1}{7x^{2}}}{1+\frac{1}{x^{4}}}$ Since $\lim_{x\to\pm\infty}\frac{1}{x^{k}} = 0$ for $k>0$, we have $\lim_{x\to\pm\infty}\frac{7}{x^{2}}\cdot\frac{1+\frac{1}{7x}+\frac{1}{7x^{2}}}{1+\frac{1}{x^{4}}}=0$.

Step2: Determine end - behavior

The end - behavior of the function $y = f(x)$ is determined by the limit as $x\to\pm\infty$. Since $\lim_{x\to\pm\infty}\frac{7x^{2}+x + 1}{x^{4}+1}=0$, as $x\to\pm\infty$, $f(x)\to0$.

Answer:

The correct statement is $\lim_{x\to\pm\infty}\frac{7x^{2}+x + 1}{x^{4}+1}=\lim_{x\to\pm\infty}\frac{7}{x^{2}}$, so the end behavior of the function is that as $x\to\pm\infty$, $f(x)\to0$.