Question

A function f(x) is defined as 

\[f\left( x \right) = \begin{cases}\frac{x^2 - 9}{x - 3}; if & x \neq 3 \\ 6 ; if & x = 3\end{cases}\]

Show that f(x) is continuous at x = 3

 

Answer 1

Given: 

\[f\left( x \right) = \binom{\frac{x^2 - 9}{x - 3}, if x \neq 3}{6, if x = 3}\] 

We observe

(LHL at x = 3) = 

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

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

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

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

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

Given

\[f\left( 3 \right) = 6\]

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

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

x = 3