\( = \lim _ { t \rightarrow \infty } [ \frac { 2 } { 3 } \ln ( 3 x + 6 ) ] _ { 1 } ^ { t } \)

To evaluate the limit \[ \lim_{t \rightarrow \infty} \left[ \frac{2}{3} \ln(3x + 6) \right]_{1}^{t}, \] we first need to understand what the expression \(\left[ \frac{2}{3} \ln(3x + 6) \right]_{1}^{t}\) means. This notation indicates that we are evaluating the function \(\frac{2}{3} \ln(3x + 6)\) at the upper limit \(t\) and the lower limit \(1\), and then taking the difference: \[ \frac{2}{3} \ln(3t + 6) - \frac{2}{3} \ln(3 \cdot 1 + 6). \] Now, we can simplify this expression: 1. Evaluate the lower limit: \[ \frac{2}{3} \ln(3 \cdot 1 + 6) = \frac{2}{3} \ln(3 + 6) = \frac{2}{3} \ln(9). \] 2. Now, evaluate the upper limit as \(t\) approaches infinity: \[ \frac{2}{3} \ln(3t + 6). \] As \(t\) approaches infinity, \(3t + 6\) also approaches infinity, and thus \(\ln(3t + 6)\) approaches infinity. Therefore, we have: \[ \lim_{t \rightarrow \infty} \frac{2}{3} \ln(3t + 6) = \infty. \] 3. Now, we can combine these results: \[ \lim_{t \rightarrow \infty} \left[ \frac{2}{3} \ln(3t + 6) - \frac{2}{3} \ln(9) \right] = \infty - \frac{2}{3} \ln(9). \] Since \(\infty - \text{(a finite number)} = \infty\), we conclude that: \[ \lim_{t \rightarrow \infty} \left[ \frac{2}{3} \ln(3x + 6) \right]_{1}^{t} = \infty. \] Thus, the final result is: \[ \boxed{\infty}. \]