The value of $\underset{x→∞}{\lim}(\frac{3x-4}{3x+2})^{\frac{x+1}{3}}$ is
Answer & explanation
Correct answer: option 1
$\underset{x→∞}{\lim}(1-\frac{6}{3x+2})^{\frac{x+1}{3}}$
$⇒e^{\underset{x→∞}{\lim}(\frac{-6}{3x+2})(\frac{x+1}{3})}⇒e^{\underset{x→∞}{\lim}\frac{-2(1+1/x)}{(3+2/x)}}=e^{-2/3}$