limn→∞ n2−n+1n2−n−1n(n−1) is equal to

=limn→∞ n2−n+1n2−n−1n(n−1) =limn→∞ n(n−1)+1n(n−1)−1n(n−1)=limn→∞ 1+1n(n−1)n(n−1)1−1n(n−1)n(n−1) =ee−1 =e2

limn→∞ n2−n+1n2−n−1n(n−1) is equal to

=limn→∞ n2−n+1n2−n−1n(n−1) =limn→∞ n(n−1)+1n(n−1)−1n(n−1)=limn→∞ 1+1n(n−1)n(n−1)1−1n(n−1)n(n−1) =ee−1 =e2

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$lim_{n \to \infty} {(\frac{n^{2} - n + 1}{n^{2} - n - 1})}^{n (n - 1)}$ is equal to

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$e^{2}$

$e^{- 1}$

answer is B.

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Detailed Solution

$= lim_{n \to \infty} {(\frac{n^{2} - n + 1}{n^{2} - n - 1})}^{n (n - 1)} = lim_{n \to \infty} {(\frac{n (n - 1) + 1}{n (n - 1) - 1})}^{n (n - 1)}$

$= lim_{n \to \infty} \frac{{(1 + \frac{1}{n (n - 1)})}^{n (n - 1)}}{{(1 - \frac{1}{n (n - 1)})}^{n (n - 1)}} = \frac{e}{e^{- 1}} = e^{2}$

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limn→∞ n2−n+1n2−n−1n(n−1) is equal to

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$lim_{n \to \infty} {(\frac{n^{2} - n + 1}{n^{2} - n - 1})}^{n (n - 1)}$ is equal to