Question

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

Answer 1

$$\begin{gathered} \underset{x\to 3}{\lim }\left[\frac{x^2-4x+3}{x^2-2x-3}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{0}{0}. \\ \underset{x\to 3}{\lim }\left[\frac{x^2-x-3x+3}{x^2-3x+x-3}\right] \\ =\underset{x\to 3}{\lim }\left[\frac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x-3\right)}\right] \\ =\underset{x\to 3}{\lim }\left(\frac{x-1}{x+1}\right) \\ =\frac{3-1}{3+1} \\ =\frac{1}{2} \end{gathered}$$