How do you find the asymptotes for f(x) = x / sqrt(x^2 + 1)?

We can rewrite this as follows

$f \left(x\right) = \frac{x}{\left\mid x \right\mid \sqrt{1 + \frac{1}{x} ^ 2}}$

Hence for $x \to + \infty$ we have that $f \left(x\right) \to 1$

and

for $x \to - \infty$ we have that $f \left(x\right) \to - 1$

Hence the horizontal asympotes are $y = 1$ and $y = - 1$