Question

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

• A. $x=0$ only
• B. $x=2$ only
• C. $x=0$ and $2$
• D. None of the above

Answer 1

The correct option is C $x=0$ and $2$

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