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Question:
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Let $f$ be the function defined by
$f(x)=\left\{\begin{array}{ll}
\frac{x^{2}-1}{x^{2}-2|x-1|-1}, & x \neq 1 \\
1 / 2, & x=1
\end{array}\right.$
Options:
$f(x)=\left\{\begin{array}{ll}
\frac{x^{2}-1}{x^{2}-2|x-1|-1}, & x \neq 1 \\
1 / 2, & x=1
\end{array}\right.$
Solution:
1928 Upvotes
Verified Answer
The correct answer is:
The function is not continuous at $x=1$
For $x < 1, f(x)=\frac{x^{2}-1}{x^{2}+2 x-3}=\frac{x+1}{x+3}$ $\therefore \lim _{x \rightarrow 1^{-}} f(x)=\frac{1}{2}$
For $x>1, f(x)=\frac{x^{2}-1}{x^{2}-2 x+1}=\frac{x+1}{x-1}$
$$
\therefore \lim _{\mathrm{x} \rightarrow 1^{+}} \mathrm{f}(\mathrm{x})=\infty
$$
$\therefore$ the function is not continuous at $x=1$.
For $x>1, f(x)=\frac{x^{2}-1}{x^{2}-2 x+1}=\frac{x+1}{x-1}$
$$
\therefore \lim _{\mathrm{x} \rightarrow 1^{+}} \mathrm{f}(\mathrm{x})=\infty
$$
$\therefore$ the function is not continuous at $x=1$.
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