Question

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

• A. $x=0$ only
• B. $x=2$ only
• C. $x=0$ and $2$
• D. none of these

Answer 1

Answer: (C)

$f\left(x\right)=\underset{n\to \infty }{\lim }\frac{{\left[{\left(x-1\right)}^2\right]}^n-1}{{\left[{\left(x-1\right)}^2\right]}^n+1}$
$=\underset{n\to \infty }{\lim }\frac{1-\frac{1}{{\left[{\left(x-1\right)}^2\right]}^n}}{1+\frac{1}{{\left[{\left(x-1\right)}^2\right]}^n}}$
$=\{\begin{array}{ll}-1,& 0\le {\left(x-1\right)}^2<1\\ 0,& (x-1{)}^2=1\\ 1,& (x-1{)}^2>1\end{array}$
$=\{\begin{array}{ll}1,& x<0\\ 0,& x=0\\ -1,& 0<x<2\\ 0,& x=2\\ 1,& x>2\end{array}$
Thus, $f(x)$ is discontinuous at $x=0,2$