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Limits and Derivatives — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsLimits and Derivatives1 Mark
MCQ
$ \lim_\limits{\text{x} \rightarrow 1}\frac{(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{n}})}{[(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{}})]^2},n \in N,$ equals:
  • A
    $^{2\text{n}}{\text{P}}$
  • ✓
    $^{2\text{n}}{\text{C}}_\text{n}$
  • C
    $(2n) !$
  • D
    None of these
✓

Answer

Correct option: B.
$^{2\text{n}}{\text{C}}_\text{n}$
$\lim_\limits{\text{x} \rightarrow 1}\frac{(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{n}})}{[(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{}})]^2}$
$\lim_\limits{\text{x} \rightarrow 1}\frac{\frac{(1-\text{x})}{(1-\text{x})}\frac{(1-\text{x}^2)}{(1-\text{x})}...\frac{(1-\text{x}^{2\text{n}})}{(1-\text{x})}}{\Big[\frac{(1-\text{x})}{(1-\text{x})}\frac{(1-\text{x}^2)}{(1-\text{x})}...\frac{(1-\text{x}^{\text{n}})}{(1-\text{x})}\Big]^2}$
$=\frac{1\times2\times3\times\ldots(2\text{n})}{(1\times2\times3\ldots \text{n})^2} = \frac{(2n)!}{\text{n}!\text{n}!}={}^{2\text{n}}\text{C}_\text{n}$
Hence, option $B$ is correct.

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