Question

Evaluate the limit : $\underset{x\to 3}{lim}\frac{x^2-4x+3}{x^2-2x-3}$

Answer 1

Given : $\underset{x\to 3}{lim}\frac{x^2-4x+3}{x^2-2x-3}becomes\frac{0}{0}$ form

$=\underset{x\to 3}{lim}\frac{x^2-4x+3}{x^2-2x-3}$

$=\underset{x\to 3}{lim}\frac{x^2-x-3x+3}{x^2-3x+x-3}$ [On Factorization]

$=\underset{x\to 3}{lim}\frac{x(x-1)-3(x-1)}{x(x-3)+1(x-3)}$

$=\underset{x\to 3}{lim}\frac{(x-3)(x-1)}{(x-3)(x+1)}[x\ne -1,3]$

$=\underset{x\to 3}{lim}\frac{(x-1)}{(x+1)}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}$

Therefore,

$\underset{x\to 3}{lim}\frac{x^2-4x+3}{x^2-2x-3}=\frac{1}{2}$