Differentiation Question 487
Question: $ \underset{x\to 1}{\mathop{\lim }},\frac{(1-x)(1-x^{2})…(1-x^{2n})}{{{{(1-x)(1-x^{2})…(1-x^{n})}}^{2}}},n\in N, $ equals
Options:
A) $ ^{2n}P_{n} $
B) $ ^{2n}C_{n} $
C) $ (2n)! $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \underset{x\to 1}{\mathop{\lim }},\frac{(1-x)(1-x^{2})…(1-x^{2n})}{{{{(1-x)(1-x^{2})…(1-x^{n})}}^{2}}} $ $ =\underset{x\to 1}{\mathop{\lim }},\frac{( \frac{1-x}{1-x} )( \frac{1-x^{2}}{1-x} )…( \frac{1-x^{2n}}{1-x} )}{{{( ( \frac{1-x}{1-x} )( \frac{1-x^{2}}{1-x} )…( \frac{1-x^{n}}{1-x} ) )}^{2}}} $ $ =\frac{1\times 2\times 3…(2n)}{{{(1\times 2\times 3…n)}^{2}}}=\frac{(2n)!}{n!n!}{{=}^{2n}}C_{n} $
BETA
