Question

The function $f\left(x\right)=\underset{n\to 0}{lim}\frac{x^{2n}-1}{x^{2n}+1}$ is identical with the function

• A. $g\left(x\right)=sgn(x-1)$
• B. $h\left(x\right)=sgn\left({\tan }^{-1}x\right)$
• C. $u\left(x\right)=sgn(|x|-1)$
• D. $v\left(x\right)=sgn\left({\cot }^{-1}x\right)$

Answer 1

Answer: (C)

$u\left(x\right)=sgn(|x|-1)$

$f\left(x\right)=\underset{n\to \infty }{lim}\frac{x^{2n}-1}{x^{2n}+1}=\underset{n\to \infty }{lim}(1-\frac{2}{x^{2n}+1})$

if $x>1$

$1-\underset{n\to \infty }{lim}\frac{2}{x^{2n}+1}=1-\frac{2}{\infty }=1-0=1$

if $-1<x<1$

$1-\underset{n\to \infty }{lim}\frac{2}{x^{2n}+1}=1-\frac{2}{{\left(\frac{\pm 1}{a}\right)}^{\infty }+1}=1-\frac{2}{0+1}$$=-1$$(x=\frac{\pm 1}{a},a>1)$

if $x<1$

$1-\underset{n\to \infty }{lim}\frac{2}{x^{2n}+1}=1-\frac{2}{-\infty }=1+0=1$

$C.u\left(x\right)=\frac{|\left|x\right|-1|}{\left|x\right|-1}$

if $x>1\Rightarrow \left|x\right|=x$

$|\left|x\right|-1|>0$

$u\left(x\right)=\frac{|x-1|}{x-1}=1$

if $0<x<1\Rightarrow \left|x\right|=x$

$|\left|x\right|-1|<0$

$u\left(x\right)=\frac{|x-1|}{x-1}=-1$

if $-1<x<0\Rightarrow \left|x\right|=-x$

$u\left(x\right)=\frac{|-x-1|}{-x-1}=-1$

if $x<-1$ $\to \left|x\right|=-x$

$u\left(x\right)=\frac{|-x-1|}{-x-1}=1|\left|x\right|-1|>0$

From above calculations both functions behave same i.e.,

$f\left(x\right)=\{\begin{array}{c}1;x>1\\ -1;-1<x<1\\ 1;x<-1\end{array}$