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  4. f (x)= displaystyle limn→∞ ((x-1)2n-1/(x-1)2n+1) is discontinuous at

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Q. $f \left(x\right)=\displaystyle\lim_{n\to\infty} \frac{\left(x-1\right)^{2n}-1}{\left(x-1\right)^{2n}+1}$ is discontinuous at

Continuity and Differentiability

Solution:

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