Determine the following limit at infinity. \[ \lim _{x \rightarrow \infty} \frac{6+7 x+6 x^{3}}{x^{3}} \]

To determine the limit \[ \lim_{x \rightarrow \infty} \frac{6 + 7x + 6x^3}{x^3}, \] we start by dividing each term in the numerator by \(x^3\): \[ \frac{6 + 7x + 6x^3}{x^3} = \frac{6}{x^3} + \frac{7x}{x^3} + \frac{6x^3}{x^3}. \] This simplifies to: \[ \frac{6}{x^3} + \frac{7}{x^2} + 6. \] Next, we evaluate the limits of each term independently as \(x \rightarrow \infty\): 1. \(\lim_{x \rightarrow \infty} \frac{6}{x^3} = 0\) because the numerator is a constant and the denominator grows without bound. 2. \(\lim_{x \rightarrow \infty} \frac{7}{x^2} = 0\) for the same reason; the numerator is constant and the denominator goes to infinity. The third term is simply a constant: \[ \lim_{x \rightarrow \infty} 6 = 6. \] Combining these limits gives: \[ \lim_{x \rightarrow \infty} \left(\frac{6}{x^3} + \frac{7}{x^2} + 6\right) = 0 + 0 + 6 = 6. \] Thus, the limit is \[ \boxed{6}. \]