Let Ltn→∞x2n-1x2n+1, then

Ltn→∞x2n-1x2n+1 x>1 =Ltn→∞ 1-1x2n1+1x2n=1 x < 1 Ltn→∞ x2n-1x2n+1=0-10+1=-1

Let Ltn→∞x2n-1x2n+1, then

Ltn→∞x2n-1x2n+1 x>1 =Ltn→∞ 1-1x2n1+1x2n=1 x < 1 Ltn→∞ x2n-1x2n+1=0-10+1=-1

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Let $\underset{n \to \infty}{Lt} \frac{x^{2 n} - 1}{x^{2 n} + 1}$ , then

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$f (x) = \{\begin{cases} 1, & for | x | > 1 \\ - 1, & for | x | \leq 1 \end{cases}$

1 f is not defined for any value of ' x '

$f (x) = 1, for |x| = 1$

$f (x) = \{\begin{matrix} 1, & for | x | > 1 \\ - 1, & for | x | < 1 \end{matrix}$

answer is B.

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Detailed Solution

$\underset{n \to \infty}{Lt} \frac{x^{2 n} - 1}{x^{2 n} + 1} |x| > 1 = \underset{n \to \infty}{Lt} \frac{1 - \frac{1}{x^{2 n}}}{1 + \frac{1}{x^{2 n}}} = 1 |x| < 1 \underset{n \to \infty}{Lt} \frac{x^{2 n} - 1}{x^{2 n} + 1} = \frac{0 - 1}{0 + 1} = - 1$