First slide

Introduction to limits

Question

$lim_{x \to - 1} \frac{(1 + x) (1 - x^{2}) (1 + x^{3}) (1 - x^{4}) \dots (1 - x^{4 n})}{{[(1 + x) (1 - x^{2}) (1 + x^{3}) (1 - x^{4}) \dots (1 - x^{2 n})]}^{2}}$ is equal to

Moderate

A

$^{4 n} C_{2 n}$
B

$^{2 n} C_{n}$
C

${2.}^{4 n} C_{2 n}$
D

${2.}^{2 n} C_{n}$

Solution

We have,
$\begin{array}{l} lim_{x \to - 1} \frac{(1 + x) (1 - x^{2}) (1 + x^{3}) (1 - x^{4}) \dots (1 + x^{4 n - 1}) (1 - x^{4 n})}{{[(1 + x) (1 - x^{2}) (1 + x^{3}) (1 - x^{4}) \dots (1 - x^{2 n})]}^{2}} \\ = lim_{x \to - 1} \frac{(1 + x^{2 n + 1}) (1 - x^{2 n + 2}) \dots (1 + x^{4 n - 1}) (1 - x^{4 n})}{(1 + x) (1 - x^{2}) (1 + x^{3}) (1 - x^{4}) \dots (1 - x^{2 n})} \\ = lim_{x \to - 1} \{\frac{1 + x^{2 n + 1}}{1 + x} \times \frac{1 - x^{2 n + 2}}{1 - x^{2}} \times \frac{1 + x^{2 n + 3}}{1 + x^{3}} \times \dots \times \frac{1 - x^{4 n}}{1 - x^{2 n}}\} \end{array}$
$\begin{array}{l} = lim_{x \to - 1} \{\frac{x^{2 n + 1} + 1}{x + 1} \times \frac{x^{2 n + 2} - 1}{x^{2} - 1} \times \frac{x^{2 n + 3} + 1}{x^{3} + 1} \times \dots \times \frac{x^{4 n} - 1}{x^{2 n} - 1}\} \\ = lim_{x \to - 1} \{\frac{x^{2 n + 1} - (- 1)^{2 n + 1}}{x - (- 1)} \times \frac{x^{2 n + 2} - (- 1)^{2 n + 2}}{x^{2} - (- 1)^{2}} \\ = \frac{2 n + 1}{1} \times \frac{x^{2 n + 3} - (- 1)^{2 n + 3}}{x^{3} - (- 1)^{3}} \times \dots \dots \times \frac{x^{4 n} - (- 1)^{4 n}}{x^{2 n} - (- 1)^{2 n}}\} \\ = \frac{4 n!}{((2 n)!}^{2}} =^{4 n} C_{2 n} . \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App