1. **State the problem:** Find the limit as $x$ approaches infinity of the function $$\frac{4x^2 + 3x + 4}{3x + 1}.$$\n\n2. **Recall the rule for limits at infinity of rational functions:** When $x \to \infty$, the behavior of a rational function is dominated by the highest degree terms in numerator and denominator.\n\n3. **Identify the highest degree terms:** Numerator highest degree term is $4x^2$, denominator highest degree term is $3x$.\n\n4. **Divide numerator and denominator by the highest power of $x$ in the denominator, which is $x$: **\n$$\frac{\frac{4x^2}{x} + \frac{3x}{x} + \frac{4}{x}}{\frac{3x}{x} + \frac{1}{x}} = \frac{4x + 3 + \frac{4}{x}}{3 + \frac{1}{x}}.$$\n\n5. **As $x \to \infty$, terms with $\frac{1}{x}$ approach 0, so simplify:**\n$$\lim_{x \to \infty} \frac{4x + 3 + 0}{3 + 0} = \lim_{x \to \infty} \frac{4x + 3}{3}.$$\n\n6. **Since $4x$ grows without bound, the limit is:**\n$$\lim_{x \to \infty} \frac{4x + 3}{3} = \infty.$$\n\n**Final answer:** The limit is $\infty$ (the function grows without bound as $x$ approaches infinity).