Solution
= Solution
Write $R(x)=(x^2+3)(2x^2-1)/(3x^4-x^3-1)$. This <rational function> is <continuous> at both finite points because its denominator is nonzero there. Substitution gives the first two <limits>; at infinity divide numerator and denominator by $x^4$ and use their leading coefficients. Therefore
$$
\boxed{\lim_{x\to0}R(x)=3,\qquad\lim_{x\to1}R(x)=4,\qquad\lim_{x\to\infty}R(x)=\frac23.}
$$