Limn -n2 - n-1n2 - n - 1 nn-1 is equal to

Lim_n →∞(n^2 n+1n^2 n 1 )^n(n 1 ) =e^lim_n →∞ (n^2 n+1n^2 n 1 1 )(n(n 1 )) = e^lim_n →∞ (2n^2 n 1 )(n(n 1 ))= e^2 ...

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Byju's Answer

Standard XII

Mathematics

Single Point Continuity

limn →∞n2 - n...

Question

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

e

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e^{2}

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e^{- 1}

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1

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Solution

The correct option is B $e^{2}$
$lim n \to \infty {(\frac{n^{2} - n + 1}{n^{2} - n - 1})}^{n (n - 1)}$
$= e^{lim n \to \infty ⎛ ⎜ ⎝ \frac{n^{2} - n + 1}{n^{2} - n - 1} - 1 ⎞ ⎟ ⎠ (n (n - 1))}$
$= e^{lim n \to \infty ⎛ ⎝ \frac{2}{n^{2} - n - 1} ⎞ ⎠ (n (n - 1))} = e^{2}$

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Single Point Continuity

Standard XII Mathematics

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limn→∞(n2−n+1n2−n−1)n(n−1) is equal to

The correct option is B e2limn→∞(n2−n+1n2−n−1)n(n−1) =elimn→∞⎛⎜⎝n2−n+1n2−n−1−1⎞⎟⎠(n(n−1)) =elimn→∞⎛⎝2n2−n−1⎞⎠(n(n−1))=e2

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

The correct option is B $e^{2}$
$lim n \to \infty {(\frac{n^{2} - n + 1}{n^{2} - n - 1})}^{n (n - 1)}$
$= e^{lim n \to \infty ⎛ ⎜ ⎝ \frac{n^{2} - n + 1}{n^{2} - n - 1} - 1 ⎞ ⎟ ⎠ (n (n - 1))}$
$= e^{lim n \to \infty ⎛ ⎝ \frac{2}{n^{2} - n - 1} ⎞ ⎠ (n (n - 1))} = e^{2}$