\[\underset{n\to \infty }{\mathop{\lim }}\,\,{{\left( \frac{{{n}^{2}}-n+1}{{{n}^{2}}-n-1} \right)}^{n(n-1)}}=\] [AMU 2002]

\[\underset{n\to \infty }{\mathop{\lim }}\,\,{{\left( \frac{{{n}^{2}}-

1.	\[\underset{n\to \infty }{\mathop{\lim }}\,\,{{\left( \frac{{{n}^{2}}-n+1}{{{n}^{2}}-n-1} \right)}^{n(n-1)}}=\] [AMU 2002]
A.	e
B.	\[{{e}^{2}}\]
C.	\[{{e}^{-1}}\]
D.	1
Answer» C. \[{{e}^{-1}}\]

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\[\underset{n\to \infty }{\mathop{\lim }}\,\,{{\left( \frac{{{n}^{2}}-n+1}{{{n}^{2}}-n-1} \right)}^{n(n-1)}}=\] [AMU 2002]

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