Big O of rational function using the definition
Rewrite $\frac{3x^3+2x^2+x+1}{4x^2+1} = \frac{3}{4}x \cdot \frac{12x^3+8x^2+4x+4}{12x^3+3x} = \frac{3}{4}x (1 + \frac{8x^2+x+4}{12x^3+3x})$.

Clearly, the limit of $f(x) = \frac{8x^2+x+4}{12x^3+3x}$ for $x \to +\infty$ is $0$ and hence, for any constant $c'&gt;0$, there is a constant $x'_0$ such that $f(x) \le c'$ for all $x \ge x'_0$. Pick your favorite constant $c'$, for example $c'=1$. You want to figure out any value of $x'_0$ for which $f(x) \le 1$ or, equivalently:

$$
12x^3+2x - 4 \ge 8x^2.
$$

This is clearly true for $x'_0=1$. Indeed, if $x \ge 1$:
$$
12x^3+2x - 4 &gt; 12x^2 - 4 \ge 8x^2 + 4x^2 - 4 \ge 8x^2.
$$

Going back to the original formula we can conclude that, for every $x \ge 1$,
$$
 \frac{3}{4} x \cdot \left(1 + \frac{8x^2+x+4}{12x^3+3x} \right)
\le \frac{3}{4} x \cdot (1 + 1)
= \frac{3}{2} x.
$$

Which satisfies the definition of $O(x)$ once you pick $c = \frac{3}{2}$ and $x_0=1$.