Question #e27b8

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Question #e27b8

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Answer 1 · 2016-10-20T22:45:23

For $x \neq 0$ $x != 0$ , we have

$\frac{x}{\sqrt{x^{2} + 1}} = \frac{x}{\sqrt{x^{2} (1 + \frac{1}{x^{2}})}}$ $x/sqrt(x^2+1) = x/sqrt(x^2(1+1/x^2))$

$= \frac{x}{\sqrt{x^{2}} \sqrt{1 + \frac{1}{x^{2}}}}$ $= x/(sqrt(x^2)sqrt(1+1/x^2))$

Since $\sqrt{x^{2}} = | x |$ $sqrt(x^2) = absx$ , we have

$lim_(xrarroo) x/sqrt(x^2+1) =lim_(xrarroo)x/(xsqrt(1+1/x^2))$

$= lim_(xrarroo)1/sqrt(1+1/x^2)$

$= 1/sqrt(1+0) = 1$

$y=1$ is a horizontal asymptote on the right.

On the left, we have $x < 0$ , so $sqrt(x^2) = -x$ and

$lim_(xrarr-oo) x/sqrt(x^2+1) =lim_(xrarroo)x/(-xsqrt(1+1/x^2))$

$= lim_(xrarroo)-1/sqrt(1+1/x^2)$

$= -1/sqrt(1+0) = -1$

$y= -1$ is a horizontal asymptote on the left.

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