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