Question

The value of $\underset{x\to \infty }{\lim }{\left(\frac{3x-4}{3x+2}\right)}^{\frac{x+1}{3}}$ is equal to :

• A. $e^{-1/3}$
• B. $e^{-2/3}$
• C. $e^{-1}$
• D. $e^{-2}$

Answer 1

Answer: (B)

Consider $\underset{x\to \infty }{\lim }{\left(\frac{3x-4}{3x+2}\right)}^{\frac{x+1}{3}}=\underset{x\to \infty }{\lim }{\left(\frac{3x+2-6}{3x+2}\right)}^{\frac{x+1}{3}}$
$=\underset{x\to \infty }{\lim }{\left(\frac{3x+2}{3x+2}-\frac{6}{3x+2}\right)}^{\frac{x+1}{3}}=\underset{x\to \infty }{\lim }{\left(1+\frac{-6}{3x+2}\right)}^{\frac{x+1}{3}}$
$=\underset{x\to \infty }{\lim }{\left({\left\{1+\frac{-6}{3x+2}\right\}}^{\frac{3x+2}{-6}}\right)}^{\frac{-6}{3x+2}\times \frac{x+1}{3}}$
$=\underset{x\to \infty }{\lim }{\left[e\right]}^{\frac{-6}{3x+2}\times \frac{x+1}{3}}\left(\because \underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x=e\right)$
$=\underset{x\to \infty }{\lim }{e}^{\frac{-2x-2}{3x+2}}=\underset{x\to \infty }{\lim }{e}^{\frac{x\left(-2-2/x\right)}{x\left(3+2/x\right)}}=\underset{x\to \infty }{\lim }{e}^{\frac{-2-2/x}{3+2/x}}=e^{-2/3}$