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 limit as x→∞

To find the limit of \( f(x)=\frac{2x}{1 - x^2} \) as \( x \) approaches infinity, we can divide both the numerator and the denominator by the highest power of \( x \) in the denominator, which is \( x^2 \).

Step2: Divide numerator and denominator by \( x^2 \)

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Step3: Evaluate the limit

As \( x \to \infty \), \( \frac{2}{x} \to 0 \) and \( \frac{1}{x^2} \to 0 \). So we have:

$$ \frac{0}{0 - 1}=0 $$

Wait, no, wait. Wait, let's re - do the division. Wait, the highest power in the denominator is \( x^2 \), numerator is \( 2x \), so when we divide numerator and denominator by \( x^2 \), numerator becomes \( \frac{2x}{x^2}=\frac{2}{x} \), denominator becomes \( \frac{1}{x^2}-\frac{x^2}{x^2}=\frac{1}{x^2}-1 \). But wait, another way: for rational functions, when the degree of the numerator (\( n = 1 \)) is less than the degree of the denominator (\( m=2 \)), the limit as \( x\to\pm\infty \) is 0? Wait, no, wait, let's check the original function again. Wait, \( f(x)=\frac{2x}{1 - x^2}=\frac{2x}{-x^2 + 1}=-\frac{2x}{x^2-1} \). As \( x\to\infty \), \( \frac{2x}{x^2-1}\approx\frac{2x}{x^2}=\frac{2}{x}\to0 \), so \( -\frac{2}{x}\to0 \)? Wait, no, that's wrong. Wait, wait, I made a mistake. Let's use the rule for rational functions: if \( f(x)=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0} \), then:

• If \( n\lt m \), \( \lim_{x\to\pm\infty}f(x) = 0 \)
• If \( n = m \), \( \lim_{x\to\pm\infty}f(x)=\frac{a_n}{b_m} \)
• If \( n\gt m \), the limit does not exist (or is \( \pm\infty \))

In our case, \( n = 1 \), \( m = 2 \), so \( n\lt m \), so the limit should be 0? But wait, let's check with another approach. Let's factor the denominator: \( 1 - x^2=(1 - x)(1 + x) \). As \( x\to\infty \), \( 1 - x^2\approx -x^2 \), and numerator \( 2x\approx2x \). So \( \frac{2x}{1 - x^2}\approx\frac{2x}{-x^2}=-\frac{2}{x}\to0 \) as \( x\to\infty \). Wait, but the options have "The graph approaches 0 as x approaches infinity" as one of the options. But wait, let's re - calculate the limit correctly.

Wait, let's take \( x = 1000 \), then \( f(1000)=\frac{2\times1000}{1-(1000)^2}=\frac{2000}{1 - 1000000}=\frac{2000}{-999999}\approx - 0.002 \), which is close to 0. As \( x \) gets larger, say \( x = 10000 \), \( f(10000)=\frac{2\times10000}{1-(10000)^2}=\frac{20000}{1 - 100000000}=\frac{20000}{-99999999}\approx - 0.0002 \), which is closer to 0. So the limit as \( x\to\infty \) of \( f(x) \) is 0.

Wait, but wait, the first option says approaches - 2, let's check the limit again. Wait, maybe I messed up the division. Let's try to divide numerator and denominator by \( x \) (the highest power in the numerator). Wait, numerator is \( 2x \), denominator is \( 1 - x^2 \). Divide numerator and denominator by \( x \):

\( \frac{2x\div x}{(1 - x^2)\div x}=\frac{2}{\frac{1}{x}-x} \)

As \( x\to\infty \), \( \frac{1}{x}\to0 \), and \( -x\to-\infty \), so the denominator \( \frac{1}{x}-x\to-\infty \), and \( \frac{2}{-\infty}\to0 \). So the limit is 0. So the graph approaches 0 as \( x \) approaches infinity.

Answer:

The graph approaches 0 as x approaches infinity.