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    I know it will have vertical asymptotes at x = 1 and x = -1 and that it will approach 0 as x \to \pm\infty . But I am not sure how to determine whether y tends to 0 from above/below the x-axis, as well as what happens for -1<x<1.
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    (Original post by FXLander)
    I know it will have vertical asymptotes at x = 1 and x = -1 and that it will approach 0 as x \to \pm\infty . But I am not sure how to determine whether y tends to 0 from above/below the x-axis, as well as what happens for -1<x<1.
    For infinities, consider whether each of the terms is +ve or -ve, for large values +ve and large values -ve.

    Between -1 and 1, plotting a few points is probably easiest. Alternatively consider what happens as x goes up to 1, what are the signs and relative sizes of the two terms - are they both bounded? Does one go off to infinity? Plus or minus?. Similarly as x goes down to -1.

    And plot the point for x=0.
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    (Original post by FXLander)
    I know it will have vertical asymptotes at x = 1 and x = -1 and that it will approach 0 as x \to \pm\infty . But I am not sure how to determine whether y tends to 0 from above/below the x-axis, as well as what happens for -1<x<1.
    If you say that y=\frac{2x}{x^2-1} then you can examine what happens as x \rightarrow -1_\pm. If it approaches from the LEFT, then x&lt;-1 and so the numerator is negative, while the denominator is positive (since x^2-1&lt;0 for x&lt;-1). So you essentially get negative/positive which is still negative. So as x \rightarrow 1_-, we know that y&lt;0

    Repeat similar analysis for x \rightarrow -1_+, and x \rightarrow 1_{\pm}

    As for what happens when x \in (-1,1), then consider the form above again, and branching from the last thing I said. We know that x^2-1&lt;0 and 2x&lt;0 for the first half of the interval, so y&gt;0 here. Then y&lt;0 for the second half. So as you go from x=-1 to x=1 your y should go from positive to negative while crossing the origin.
 
 
 
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