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f(x) = x/x^2-1
To identify the asymptotes.
Denominator x^2-1 = (x-1)(x+1)
For x > 0, f(x) = x/(x^2-1) > 0 and Lt x/x^2-1 = 0 as x--> 0.
Similarly , when x< -1, x/(x^2-1) < 0 as numerator is negative and denominator is positive. Therefore Lt x/(x^2-1) as x--> - infinity
Therefore x axis is an asymptote.
When x = 1, f(x) does not exist. But Right Lt x/(x^2-1) = infinity as x --> 1+
Left Lt x/(x^2-1) = -infinity as x --> 1-
Therefore x =1 is a vertical asyptote.
At x =-1 the f(x) = x /(x^2-1 doenot exist as Rt lt x/(x^2-1) is infinity and left lt x/(x^2-1 is - infinity. Thus x= -1 is another vertical asymptote.
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