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Question: \[\lim_{n \rightarrow \infty}\left( \frac{n^{2} - n + 1}{n^{2} - n - 1} \right)^{n(n - 1)} =\]...

lim⁡n→∞(n2−n+1n2−n−1)n(n−1)=\lim_{n \rightarrow \infty}\left( \frac{n^{2} - n + 1}{n^{2} - n - 1} \right)^{n(n - 1)} =

A

e

B

e2e^{2}

C

e−1e^{- 1}

D

1

Answer

e2e^{2}

Explanation

Solution

lim⁡n→∞(n2−n+1n2−n−1)n(n−1)=lim⁡n→∞(n(n−1)+1n(n−1)−1)n(n−1)\lim_{n \rightarrow \infty}\left( \frac{n^{2} - n + 1}{n^{2} - n - 1} \right)^{n(n - 1)} = \lim_{n \rightarrow \infty}\left( \frac{n(n - 1) + 1}{n(n - 1) - 1} \right)^{n(n - 1)}

=lim⁡n→∞(1+1n(n−1))n(n−1)(1−1n(n−1))n(n−1)=ee−1=e2= \lim_{n \rightarrow \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}.

Alternative Method: lim⁡n→∞(1+2n2−n−1)n(n−1)\lim_{n \rightarrow \infty}\left( 1 + \frac{2}{n^{2} - n - 1} \right)^{n(n - 1)}= elim⁡n→∞2n(n−1)n2−n−1=e2e^{\lim_{n \rightarrow \infty}\frac{2n(n - 1)}{n^{2} - n - 1}} = e^{2}.