Question

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

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

Answer 1

Answer: (A)

$e^{-2/3}$

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

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

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

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

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

It is of the form $\frac{0}{0}$, so applying L-Hospital's rule

$\Rightarrow k=e^{\underset{x\to \infty }{lim}\frac{\frac{6}{(3x+4)(3x+2)}}{\frac{-1}{{(x+1)}^2}}}$

$\Rightarrow k=e^{\frac{\frac{6}{(3+0)(3+0)}}{\frac{-1}{(0+1{)}^2}}}$

$\Rightarrow k=e^{-2/3}$