Question

Let $f$ be the function defined by $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{x^2-1}{x^2-2|x-1|-1},& x\ne 1\end{array}\\ \begin{array}{cc}\frac{1}{2},& x=1\end{array}\end{array}$ then

• A. the function is continuous for all value of $x$
• B. the function is continuous only for $x>1$
• C. the function is continuous at $x=1$
• D. the function is not continuous at $x=1$

Answer 1

Answer: (D)

the function is not continuous at $x=1$

For $x<1,f\left(x\right)=\frac{x^2-1}{x^2+2x+-3}=\frac{x+1}{x+3}$

$\therefore \underset{x\to 1^{-}}{lim}f\left(x\right)=\frac{1}{2}$

For $x>1,f\left(x\right)=\frac{x^2-1}{x^2-2x+1}=\frac{x+1}{x-1}$

$\therefore \underset{x\to 1^{-}}{lim}f\left(x\right)=\infty$

$\therefore$ The function is not continuous at $x=1.$