Question

A function f(x) is defined as,

\[f\left( x \right) = \begin{cases}\frac{x^2 - x - 6}{x - 3}; if & x \neq 3 \\ 5 ; if & x = 3\end{cases}\]  Show that f(x) is continuous that x = 3.

Answer 1

Given: 

\[f\left( x \right) = \binom{\frac{x^2 - x - 6}{x - 3}, x \neq 3}{5, x = 3}\]

We observe

(LHL at x = 3) = 

\[\lim_{x \to 3^-} f\left( x \right) = \lim_{h \to 0} f\left( 3 - h \right)\]

\[\lim_{h \to 0} \frac{\left( 3 - h \right)^2 - \left( 3 - h \right) - 6}{\left( 3 - h \right) - 3} = \lim_{h \to 0} \frac{9 + h^2 - 6h - 3 + h - 6}{- h} = \lim_{h \to 0} \frac{h^2 - 5h}{- h} = \lim_{h \to 0} \left( 5 - h \right) = 5\]

And, (RHL at x = 3)​ = 

\[\lim_{x \to 3^+} f\left( x \right) = \lim_{h \to 0} f\left( 3 + h \right)\]

\[\lim_{h \to 0} \frac{\left( 3 + h \right)^2 - \left( 3 + h \right) - 6}{\left( 3 + h \right) - 3} = \lim_{h \to 0} \frac{9 + h^2 + 6h - 3 - h - 6}{h} = \lim_{h \to 0} \frac{h^2 + 5h}{h} = \lim_{h \to 0} \left( 5 + h \right) = 5\]

Also, 

\[f\left( 3 \right) = 5\]

\[\therefore \lim_{x \to 3^+} f\left( x \right) = \lim_{x \to 3^-} f\left( x \right) = f\left( 3 \right)\]

Hence, 

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

\[x = 3\]