Which of the following rational functions behaves like $y=0$ as $x \rightarrow \pm \infty$ ? $f(x)=1-x^{2}$ $f(x)=\frac{x^{5}-1}{x^{2}+1}$ $f(x)=\frac{x^{2}+1}{x^{5}-1}$ $f(x)=\frac{x-1}{2 x+1}$ All of the above.
Solution
Step 1 ：Let's check each function one by one.
Step 2 ：For the function \(f(x)=1-x^{2}\), the degree of the numerator is 2, which is greater than the degree of the denominator (which is 0), so this function does not behave like \(y=0\) as \(x \rightarrow \pm \infty\).
Step 3 ：For the function \(f(x)=\frac{x^{5}-1}{x^{2}+1}\), the degree of the numerator is 5, which is greater than the degree of the denominator (which is 2), so this function does not behave like \(y=0\) as \(x \rightarrow \pm \infty\).
Step 4 ：For the function \(f(x)=\frac{x^{2}+1}{x^{5}-1}\), the degree of the numerator is 2, which is less than the degree of the denominator (which is 5), so this function behaves like \(y=0\) as \(x \rightarrow \pm \infty\).
Step 5 ：For the function \(f(x)=\frac{x-1}{2 x+1}\), the degree of the numerator is 1, which is equal to the degree of the denominator (which is also 1), so this function does not behave like \(y=0\) as \(x \rightarrow \pm \infty\).
Step 6 ：Therefore, the only function that behaves like \(y=0\) as \(x \rightarrow \pm \infty\) is \(f(x)=\frac{x^{2}+1}{x^{5}-1}\).
Step 7 ：Final Answer: \(\boxed{f(x)=\frac{x^{2}+1}{x^{5}-1}}\)
