Answer
$$\eqalign{
& {\text{No vertical asymptotes}} \cr
& {\text{Horizontal asymptotes: }}y = - 3{\text{ and }}y = 3 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{3x}}{{\sqrt {{x^2} + 2} }} \cr
& {\text{The function is continuous for all real number }}x \cr
& {\text{there are no vertical asymptotes}}{\text{.}} \cr
& {\text{Calculate the horizontal asymptotes}}{\text{, Evaluating }} \cr
& \mathop {\lim }\limits_{x \to - \infty } f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to \infty } f\left( x \right) \cr
& {\text{For }}x < 0,{\text{ we can write }}x = - \sqrt {{x^2}} ,{\text{ then dividing the}} \cr
& {\text{ numerator and denominator by }}x{\text{ produces}} \cr
& *\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{3x}}{{\sqrt {{x^2} + 2} }} = \frac{{\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{3x}}{x}} \right)}}{{\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{\sqrt {{x^2} + 2} }}{{ - \sqrt {{x^2}} }}} \right)}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( 3 \right)}}{{ - \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {1 + \frac{2}{{{x^2}}}} } \right)}} = \frac{3}{{ - \sqrt {1 + 0} }} = - 3 \cr
& {\text{For }}x > 0,{\text{ we can write }}x = \sqrt {{x^2}} ,{\text{ then dividing the }} \cr
& {\text{numerator and denominator by }}x{\text{ produces}} \cr
& *\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{3x}}{x}}}{{\frac{{\sqrt {{x^2} + 2} }}{{\sqrt {{x^2}} }}}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{3x}}{x}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {\frac{{{x^2}}}{{{x^2}}} + \frac{2}{{{x^2}}}} } \right)}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( 3 \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {1 + \frac{2}{{{x^2}}}} } \right)}} = \frac{3}{{\sqrt {1 + 0} }} = 3 \cr
& \cr
& {\text{Therefore}}{\text{, }} \cr
& {\text{No vertical asymptotes}} \cr
& {\text{Horizontal asymptotes: }}y = - 3{\text{ and }}y = 3 \cr} $$
