Question

$y=f\left(x\right)=\frac{x^2-3x+2}{x^2+x-6}.$

Find the limit of f(x) as x approaches 2.

Answer 1

WE have $f\left(x\right)=\frac{x^2-3x+2}{x^2+x-6}=\frac{(x-1)(x-2)}{(x+3)(x-2)}$ The function f is not defined at x=2 and x=-3.Hence domain of $f=\{x:xϵR,x\ne -3,x\ne 2\}.$
$\underset{x\to 2}{lim}\frac{x^2-3x+2}{x^2+x-6}=\underset{x\to 2}{lim}\frac{(x-2)(x-1)}{(x-2)(x+3)}$
$=\underset{x\to 2}{lim}\frac{x-1}{x+3}=\frac{2-1}{2+3}=\frac{1}{5}.$ To find the range of f we first observe that it cannot take the value $\frac{1}{5}$ since it is not defined at x=2.Also for $x\ne 2$, we have $y=f\left(x\right)=\frac{x-1}{x+3}$ or xy+3y=x-1 or $x=-\frac{3y+1}{y-1}.$ Hence $y\ne 1$ in the domain of x. Also at $x=-3,y=\infty .$Thus f(x) takes all real values in the domain of x except $y=\frac{1}{5}$ and y=1.Hence range of $f=\{y:yϵR,y\ne 1/5,y\ne 1\}$.