Differentiation Question 34
Question: $ \underset{n\to \infty }{\mathop{\lim }}{{( \frac{n^{2}-n+1}{n^{2}-n-1} )}^{n(n-1)}} $ is equal to
Options:
A) $ e $
B) $ e^{2} $
C) $ {e^{-1}} $
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \underset{n\to \infty }{\mathop{\lim }}{{( \frac{n^{2}-n+1}{n^{2}-n-1} )}^{n(n-1)}}=\underset{n\to \infty }{\mathop{\lim }}{{( \frac{n(n-1)+1}{n(n-1)-1} )}^{n(n-1)}} $
$ =\underset{n\to \infty }{\mathop{\lim }}\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} $
BETA
