Question

$\underset{x\to \infty }{lim}{\left[\frac{3x-4}{3x+2}\right]}^{\frac{x+1}{3}}$

Answer 1

We have,

${lim}_{x\to \infty }{\left[\frac{3x-4}{3x+2}\right]}^{\frac{x+1}{3}}$

${lim}_{x\to \infty }{[1+\frac{(-6)}{3x+2}]}^{\frac{x+1}{3}}$

We know that

${lim}_{x\to \infty }{(1+\frac{a}{x})}^x=e^a$

Therefore,

$=e^{-6}$

$=\frac{1}{e^6}$

Hence, the value is $\frac{1}{e^6}$.