A) \[x=0\]only
B) \[x=2\]only
C) \[x=0\]and 2
D) none of these
Correct Answer: C
Solution :
[c] \[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{[{{(x-1)}^{2}}]}^{n}}-1}{{{[{{(x-1)}^{2}}]}^{n}}+1}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1-\frac{1}{{{[{{(x-1)}^{2}}]}^{n}}}}{1+\frac{1}{{{[{{(x-1)}^{2}}]}^{n}}}}\] \[=\left\{ \begin{matrix} -1,\,\,\,\,0\le {{(x-1)}^{2}}<1 \\ 0,\,\,\,\,{{(x-1)}^{2}}=1 \\ 1,\,\,\,\,\,{{(x-1)}^{2}}>1 \\ \end{matrix} \right.\] \[=\left\{ \begin{matrix} 1,\,\,\,\,\,\,x<0 \\ \begin{align} & 0,\,\,\,\,\,\,\,\,\,x=0 \\ & -1,\,\,\,\,\,0<x<2 \\ \end{align} \\ 0,\,\,\,\,\,\,x=2 \\ 1,\,\,\,\,\,\,x>2 \\ \end{matrix} \right.\] Thus, f(x) is discontinuous at x=0.2.
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