Answer
$$\lim _{x\rightarrow \infty }\dfrac {2x+1}{3x^{4}-2}=\lim _{x\rightarrow -\infty }\dfrac {2x+1}{3x^{4}-2}=y=0$$
Work Step by Step
$$\lim _{x\rightarrow \infty }\dfrac {2x+1}{3x^{4}-2}=\dfrac {\dfrac {2}{x^{3}}+\dfrac {1}{x^{4}}}{3-\dfrac {2}{x^{4}}}=\dfrac {0+0}{3-0}=\dfrac {0}{3}=0$$
$$\lim _{x\rightarrow -\infty }\dfrac {2x+1}{3x^{4}-2}=\dfrac {\dfrac {2}{x^{3}}+\dfrac {1}{x^{4}}}{3-\dfrac {2}{x^{4}}}=\dfrac {0+0}{3-0}=\dfrac {0}{3}=0$$
So the horizontal asymptote will be $y=0$
