_ n , ( n^2 - n + 1 n^2 - n - 1 )^ n(n - 1) = - Vidyadip

MCQ

$\mathop {\lim }\limits_{n \to \infty } \,{\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} \right)^{n(n - 1)}} = $

A
$e$
✓
${e^2}$
C
${e^{ - 1}}$
D
$1$

✓

Answer

Correct option: B.

${e^2}$

b
(b) $\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} \right)^{n(n - 1)}} = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{n(n - 1) + 1}}{{n(n - 1) - 1}}} \right)^{n(n - 1)}}$

$ = \mathop {\lim }\limits_{n \to \infty } \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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