How do you find the horizontal asymptotes of f(x)=(1-2x)/(1+x^2)^1/2
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How do you find the horizontal asymptotes of f(x)=(1-2x)/(1+x^2)^1/2

[From: ] [author: ] [Date: 12-12-23] [Hit: ]
Note: All I did was divide the numerator by the largest x value in the denominate (square root of x^2 is simply equal to x when x >/= 0). This concept applies for all limits of an equation where x approaches +/- infinity. Additionally, recall that 1/∞ = 0.Apply the same rules for as x approaches negative infinity, and you should find that y = 2.......
Assuming that the denominator is all under the square root (and not the entire function given), the horizontal asymptotes are y = -2 and y = 2.

As x approaches positive infinity,

limit f(x) = [(1-2x)/x] / [(1+x^2)/x^2]^1/2 = [1/x -2] / [((1/x^2) + 1]^1/2 = -2
x-->∞

Note: All I did was divide the numerator by the largest x value in the denominate (square root of x^2 is simply equal to x when x >/= 0). This concept applies for all limits of an equation where x approaches +/- infinity. Additionally, recall that 1/∞ = 0.

Apply the same rules for as x approaches negative infinity, and you should find that y = 2.
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