# How do you find the horizontal asymptote for f(x)=(1-2x)/sqrt(1+x^2)?

Dec 6, 2015

The horizontal asymptotes are $y = - 2$ if $x \setminus \to \setminus \infty$ and $y = 2$ if $x \setminus \to - \setminus \infty$

#### Explanation:

Horizontal asymptotes, if exist, are given by

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right)$

So, to compute your limit, observe that

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

where the last step is due to the fact that $\sqrt{{x}^{2}} = \left\mid x \right\mid$

Now, when $x \setminus \to \setminus \pm \setminus \infty$, we have that both $\frac{1}{x}$ and $\frac{1}{x} ^ 2$ tend to zero. On the other hand, $\left\mid x \right\mid = - x$ if $x \setminus \to - \setminus \infty$, and $\left\mid x \right\mid = x$ if $x \setminus \to \setminus \infty$. So, the limit becomes

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{x \left(- 2 + \cancel{\frac{1}{x}}\right)}{\left\mid x \right\mid \sqrt{\left(1 + \cancel{\frac{1}{x} ^ 2}\right)}} = \setminus \pm 1 \cdot - \frac{2}{\sqrt{1}} = - \left(\setminus \pm 2\right)$

This means that horizontal asymptotes are $y = - 2$ if $x \setminus \to \setminus \infty$ and $y = 2$ if $x \setminus \to - \setminus \infty$