Let f(x)=limn→∞ x2n−1x2n+1 Then - Infinity Learn by Sri Chaitanya

A
$f (x) = \{\begin{aligned} 1, | x | > 1 \\ - 1, | x | < 1 \end{aligned}$
B
$f (x) = \{\begin{array}{r} 1, | x | < 1 \\ - 1, | x | > 1 \end{array}$
C
$f (x)$ is not defined for any value of $x$
D
$f (x) = 1$ for $| x | = 1$

Solution:

We have,

$\begin{array}{l} lim_{n \to \infty} x^{2 n} = \{\begin{cases} 0, & if | x | < 1 \\ 1, & if | x | = 1 \\ \infty, & if | x | > 1 \end{cases} \\ ∴ f (x) = lim_{n \to \infty} \frac{1 - \frac{1}{x^{2 n}}}{1 + \frac{1}{x^{2 n}}} = lim_{n \to \infty} \{\begin{cases} - 1, & | x | < 1 \\ 0 & , | x | = 1 \\ 1 & , | x | > 1 \end{cases} \end{array}$

Let $f (x) = lim_{n \to \infty} \frac{x^{2 n} - 1}{x^{2 n} + 1}$ Then

Solution:

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