The statement that describes the behavior of the function f(x) = 2x/(1-x²) is given by: Option B: The graph approaches 0 as x approaches infinity
If limits exist, we can take limits of the function, where x tends to -∞ or ∞, and that limiting value will be the value the function will approach.
For the considered case, the function is:
[tex]f(x) = \dfrac{2x}{1-x^2}[/tex]
The missing options are:
So we need to find the limit of the function as x approaches infinity.
[tex]lim_{x\rightarrow \infty}f(x) = lim_{x\rightarrow \infty}\dfrac{2x}{1-x^2} = lim_{x\rightarrow \infty}\dfrac{2x/x^2}{1/x^2 - 1} = lim_{x\rightarrow \infty}\dfrac{2/x}{1/x^2 - 1}\\\\lim_{x\rightarrow \infty}f(x) = \dfrac{lim_{x\rightarrow \infty}(2/x^2)}{lim_{x\rightarrow \infty}(1/x^2 - 1)} = \dfrac{0}{0-1} = \dfrac{0}{-1} = 0[/tex]
Since the value of the function approaches 0 as x approaches infinity, so as its graph will do.
Thus, the statement that describes the behavior of the function f(x) = 2x/(1-x²) is given by: Option B: The graph approaches 0 as x approaches infinity
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