Limits Continuity And Differentiability Question 82
Question: $ f(x)=\underset{n\to \infty }{\mathop{\lim }}\frac{{{(x-1)}^{2n}}-1}{{{(x-1)}^{2n}}+1} $ is discontinuous at
Options:
A) $ x=0 $ only
B) $ x=2 $ only
C) $ x=0 $ and 2
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)=\underset{n\to \infty }{\mathop{\lim }}\frac{{{[{{(x-1)}^{2}}]}^{n}}-1}{{{[{{(x-1)}^{2}}]}^{n}}+1} $
$ =\underset{n\to \infty }{\mathop{\lim }}\frac{1-\frac{1}{{{[{{(x-1)}^{2}}]}^{n}}}}{1+\frac{1}{{{[{{(x-1)}^{2}}]}^{n}}}} $
$ = \begin{cases} -1,0\le {{(x-1)}^{2}}<1 \\ 0,{{(x-1)}^{2}}=1 \\ 1,{{(x-1)}^{2}}>1 \\ \end{cases} . $
$ = \begin{cases} 1,x<0 \\ \begin{matrix} & 0,x=0 \\ & -1,0<x<2 \\ \end{matrix} \\ 0,x=2 \\ 1,x>2 \\ \end{cases} . $
Thus, f(x) is discontinuous at x=0.2.
BETA
