Помогите пожалуйста решить

$\lim_{n \to \infty} ( \frac{6n-7}{6n+4} )^{3n+2}=\{( \frac{ \infty}{ \infty} )^{ \infty} \}=\{(\frac{6}{6})^{ \infty}) \}=\{1^ \infty\}= \\ \\$

делим столбиком (6n-7) на (6n+4) и получаем (6n-7)(6n+4)=1+ (-11 / 6n+4)

$= \lim_{n \to \infty} (1+ \frac{-11}{6n+4})^{3n+2}= \lim_{n \to \infty} (1+ \frac{-11}{6n+4})^{(3n+2)* \frac{6n+4}{-11} * \frac{-11}{6n+4} }= \\ \\ =\lim_{n \to \infty} ((1+ \frac{-11}{6n+4})^{\frac{6n+4} {-11}})^{ \frac{-11(3n+2) }{6n+4} }}= \\ \\=\lim_{n \to \infty} ((1+ \frac{-11}{6n+4})^{\frac{6n+4} {-11}})^{ \frac{-11(3n+2) }{2(3n+2)} }}=\lim_{n \to \infty} ((1+ \frac{-11}{6n+4})^{\frac{6n+4} {-11}})^{ \frac{-11}{2}} = \\ \\ =e^{ \frac{-11}{2} }=e^{-5.5}$