Explanation

(b)
${\lim_{x \rightarrow \infty}\left( \frac{3x - 4}{3x + 2} \right)}^{\frac{x + 1}{3}}$
$= {\lim_{x \rightarrow \infty}\left\lbrack 1 + \frac{- 6}{3x + 2} \right\rbrack}^{\frac{x + 1}{3}}$
$= \left\lbrack {\lim_{x \rightarrow \infty}\left\{ 1 + \frac{- 6}{3x + 2} \right\}}^{\frac{3x + 2}{- 6}} \right\rbrack^{\frac{- 6}{3x + 2} \times \frac{x + 1}{3}}$
$= \lbrack e\rbrack^{\lim_{x \rightarrow \infty}\ \frac{- 6}{3x + 2} \times \frac{x + 1}{3}\ }$ $\left\lbrack \because\ \lim_{x \rightarrow \infty}\left( 1 + \frac{1}{x} \right)^{x} = e\ \right\rbrack$
$= e^{\lim_{x \rightarrow \infty}\frac{- 2x - 2}{3x + 2}} = e^{- 2/3}$