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