Question

If \[f\left( x \right) = \begin{cases}\frac{x^2 - 1}{x - 1}; for & x \neq 1 \\ 2 ; for & x = 1\end{cases}\] Find whether f(x) is continuous at x = 1.

 

Answer 1

Given:

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

We observe
(LHL at x = 1) = 

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

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

(RHL at x = 1) = 

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

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

Given: 

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

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

Hence

\[f\left( x \right)\]  is continuous at 

\[x = 1\]