Question:

<p><strong>Step 1:</strong> Recognize this is of the form 1^∞. Rewrite the fraction:</p><p>$$\frac{3x-4}{3x+2} = \frac{3x+2-6}{3x+2} = 1 + \frac{-6}{3x+2}$$</p><p><strong>Step 2:</strong> Rewrite the limit using the standard form lim(1 + u)^n = e^(lim u·n):</p><p>$$\lim_{x \to \infty} \left(1 + \frac{-6}{3x+2}\right)^{\frac{x+1}{3}}$$</p><p><strong>Step 3:</strong> Apply the exponential limit formula. The exponent in the exponential is:</p><p>$$\lim_{x \to \infty} \frac{-6}{3x+2} \cdot \frac{x+1}{3} = \lim_{x \to \infty} \frac{-6(x+1)}{3(3x+2)}$$</p><p><strong>Step 4:</strong> Evaluate the limit by dividing numerator and denominator by x:</p><p>$$= \lim_{x \to \infty} \frac{-6(1 + 1/x)}{3(3 + 2/x)} = \frac{-6(1)}{3(3)} = \frac{-6}{9} = -\frac{2}{3}$$</p><p><strong>Step 5:</strong> Therefore:</p><p>$$\lim_{x \to \infty} \left(\frac{3x-4}{3x+2}\right)^{\frac{x+1}{3}} = e^{-2/3}$$</p>