A) \[{{e}^{-1/3}}\]
B) \[{{e}^{-2/3}}\]
C) \[{{e}^{-1}}\]
D) \[{{e}^{-2}}\]
Correct Answer: B
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{3x-4}{3x+2} \right)}^{\frac{x+1}{3}}}=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{3x+2-6}{3x+2} \right)}^{\frac{x+1}{3}}}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1-\frac{6}{3x+2} \right)}^{\frac{x+1}{3}}}=\underset{x\to \infty }{\mathop{\lim }}\,{{\left[ {{\left( 1-\frac{6}{3x+2} \right)}^{\frac{3x+2}{-6}}} \right]}^{\frac{-6}{3x+2}.\frac{x+1}{3}}}\]\[=\underset{x\to \infty }{\mathop{\lim }}\,\,\,{{e}^{\frac{-2\left( x+1 \right)}{3x+2}}}={{e}^{-2/3}}\]
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