Question

Discuss the continuity of the following functions at the indicated point(s): 

\[f\left( x \right) = \begin{cases}\frac{\left| x^2 - 1 \right|}{x - 1}, for & x \neq 1 \\ 2 , for & x = 1\end{cases}at x = 1\]

Answer 1

Given : 

\[f\left( x \right) = \binom{\frac{\left| x^2 - 1 \right|}{x - 1}, x \neq 1}{2, x = 1}\]

\[\Rightarrow f\left( x \right) = \begin{cases}x + 1, x < - 1 \\ \begin{array}- x - 1, - 1 \leq x < 1 \\ x + 1, x > 1\end{array} \\ 2, x = 1\end{cases}\] 

We observe

(LHL at x = 1) = 

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right) = \lim_{h \to 0} - \left( 1 - h \right) - 1 = \lim_{h \to 0} - 2 + h = - 2\]

\[f\left( 1 \right) = 2\]

\[\Rightarrow \lim_{x \to 1^-} f\left( x \right) \neq f\left( 1 \right)\]

Hence, f(x) is discontinuous at x = 1.