Question

which statement describes the behavior of the function $f(x) = \frac{2x}{1 - x^2}$? the graph approaches $-2$ as $x$ approaches infinity. the graph approaches $0$ as $x$ approaches infinity. the graph approaches $1$ as $x$ approaches infinity. the graph approaches $2$ as $x$ approaches infinity.

Explanation:

Step1: Analyze the function for end - behavior

To find the end - behavior of the rational function \(f(x)=\frac{2x}{1 - x^{2}}\) as \(x\to\infty\), we can divide both the numerator and the denominator by the highest power of \(x\) in the denominator. The highest power of \(x\) in the denominator \(1 - x^{2}\) is \(x^{2}\).

So, we rewrite the function as:

$$
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$$

Step2: Evaluate the limit as \(x\to\infty\)

As \(x\to\infty\), we know that \(\lim_{x\to\infty}\frac{2}{x} = 0\) (because the denominator \(x\) becomes very large, so the fraction \(\frac{2}{x}\) approaches \(0\)) and \(\lim_{x\to\infty}\frac{1}{x^{2}}=0\) (similarly, the denominator \(x^{2}\) becomes very large, so the fraction \(\frac{1}{x^{2}}\) approaches \(0\)).

Now, we find \(\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{\frac{2}{x}}{\frac{1}{x^{2}}-1}\)

Substitute the limits of the numerator and the denominator:
The numerator \(\lim_{x\to\infty}\frac{2}{x} = 0\) and the denominator \(\lim_{x\to\infty}(\frac{1}{x^{2}}-1)=0 - 1=-1\)

So, \(\lim_{x\to\infty}f(x)=\frac{0}{-1}=0\)

Answer:

The graph approaches \(0\) as \(x\) approaches infinity.