The function f (x) = Lim _ n , , x^ 2n - 1 x^ 2n + 1 is identical… - Vidyadip

MCQ

The function $f (x) =$ $\mathop {Lim}\limits_{n \to \infty } \,\,\frac{{{x^{2n}} - 1}}{{{x^{2n}} + 1}}$ is identical with the function

A
$g (x) = sgn(x - 1)$
B
$h (x) = sgn (tan^{-1}x)$
✓
$u (x) = sgn( | x | - 1)$
D
$v (x) = sgn (cot^{-1}x)$

✓

Answer

Correct option: C.

$u (x) = sgn( | x | - 1)$

c
$f (x) =$$\left[ \begin{gathered} \hfill \\ \hfill \\ \hfill \\ \hfill \\ \end{gathered} \right.$ $\begin{gathered} - 1\,\,\,\,\,\,if\,\, - 1 < x < 1 \hfill \\ \hfill \\ \,\,\,0\,\,\,\,\,\,if\,\,x = 1\,\,or\,\, - 1 \hfill \\ \hfill \\ \,\,\,1\,\,\,\,\,\,if\,\,|x| > 1 \hfill \\ \end{gathered} $ and $g (x) = sgn(|x| - 1)$ is same

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