Question

$\underset{x\to 4}{\lim }\frac{x^2-7x+12}{x^2-3x-4}$

Answer 1

$$\begin{gathered} \underset{x\to 4}{\lim }\left[\frac{x^2-7x+12}{x^2-3x-4}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{0}{0}. \\ \underset{x\to 4}{\lim }\left[\frac{x^2-3x-4x+12}{x^2-4x+x-4}\right] \\ =\underset{x\to 4}{\lim }\left[\frac{x\left(x-3\right)-4\left(x-3\right)}{x\left(x-4\right)+1\left(x-4\right)}\right] \\ =\underset{x\to 4}{\lim }\left[\frac{\left(x-4\right)\left(x-3\right)}{\left(x-4\right)\left(x+1\right)}\right] \\ =\frac{4-3}{4+1} \\ =\frac{1}{5} \end{gathered}$$