The function f(x) = lim_{n to infty} frac{x^{ | vedclass

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(C) To evaluate $f(x) = \lim_{n 	o \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$:Case $1$: If $|x| < 1$,then $x^{2n} 	o 0$ as $n 	o \infty$. Thus,$f(x) =

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$u(x) = 	ext{sgn}(|x| - 1)$

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(C) To evaluate $f(x) = \lim_{n 	o \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$:Case $1$: If $|x| Case $2$: If $|x| > 1$,then divide numerator and denominator by $x^{2n}$,so $f(x) = \lim_{n 	o \infty} \frac{1 - 1/x^{2n}}{1 + 1/x^{2n}} = \frac{1 - 0}{1 + 0} = 1$.Case $3$: If $x = 1$,$f(1) = \frac{1 - 1}{1 + 1} = 0$. If $x = -1$,$f(-1) = \frac{1 - 1}{1 + 1} = 0$.Thus,$f(x) = -1$ for $|x|  1$.Now,consider $u(x) = 	ext{sgn}(|x| - 1)$:If $|x| If $|x| = 1$,$|x| - 1 = 0$,so $	ext{sgn}(|x| - 1) = 0$.If $|x| > 1$,$|x| - 1 > 0$,so $	ext{sgn}(|x| - 1) = 1$.Therefore,$f(x) = u(x)$.

The function $f(x) = \lim_{n \to \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$ is identical to which of the following functions?

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