A) \[{{e}^{12}}\]
B) \[{{e}^{-12}}\]
C) \[{{e}^{4}}\]
D) \[{{e}^{3}}\]
Correct Answer: B
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1-\frac{4}{x-1} \right)}^{3x-1}}=\underset{x\to \infty }{\mathop{\lim }}\,{{\left[ {{\left( 1+\frac{(-4)}{x-1} \right)}^{\left( \frac{x-1}{-4} \right)}} \right]}^{\left( \frac{-4}{x-1} \right)(3x-1)}}\] \[={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{-4\left( 3-\frac{1}{x} \right)}{\left( 1-\frac{1}{x} \right)} \right]}}={{e}^{-12}}\].
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