Question

Solve the following:

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

• A. $e^{-\frac{2}{3}}$
• B. $e^{\frac{3}{2}}$
• C. $e^{\frac{2}{3}}$
• D. $e$

Answer 1

The correct option is A $e^{-\frac{2}{3}}$

Given limit is $\underset{x⟶\infty }{lim}{\left(\frac{3x-4}{3x+2}\right)}^{\frac{x+1}{3}}$

It is in the form of $1^{\infty }$ form,

We know that $\underset{x⟶\infty }{lim}{f\left(x\right)}^{g\left(x\right)}=e^{(f\left(x\right)-1)g\left(x\right)}$ when the limit is in $1^{\infty }$ form,

$\therefore$ the given limit is

$\underset{x⟶\infty }{lim}{\left(\frac{3x-4}{3x+2}\right)}^{\frac{x+1}{3}}$,

$=e^{\underset{x⟶\infty }{lim}\left(\frac{3x-4-3x-2}{3x+2}\right)\left(\frac{x+1}{3}\right)}$

$=e^{\underset{x⟶\infty }{lim}\left(\frac{-2}{3+\frac{2}{x}}\right)(1+\frac{1}{x})}$

$=e^{-\frac{2}{3}}$