Question

Show that the function
$f\left(x\right)=\frac{x^2-7x+12}{x-3},x\ne 3$
$-1,x=3$

is continous at $x=3$

Answer 1

$\underset{x\to 3\pm }{lim}f\left(x\right)=\underset{x\to 3^{+}}{lim}\frac{x^2-7x+12}{x-3}=\underset{x\to 3^{+}}{lim}\frac{x^2-4x-3x+12}{x-3}$

$=\underset{x\to 3\pm }{lim}\frac{x(x-4)-3(x-4)}{x-3}$

$=\underset{x\to 3\pm }{lim}\frac{(x-3)(x-4)}{(x-3)}$

$=3-4=-1$

$\because f\left(3\right)=\underset{x\to 3\pm }{lim}f\left(x\right)=-1$

i.e $f\left(x\right)$ is continuous at $x=3$