Question

помогите пожалуйста решить

Answer 1

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

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

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