Question

$\underset{x\to 3}{\lim }\frac{x^2-x-6}{x^3-3x^2+x-3}$

Answer 1

$$\begin{gathered} \underset{x\to 3}{\lim }\left[\frac{x^2-x-6}{x^3-3x^2+x-3}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{0}{0}. \\ \underset{x\to 3}{\lim }\left[\frac{x^2-3x+2x-6}{x^2\left(x-3\right)+1\left(x-3\right)}\right] \\ =\underset{x\to 3}{\lim }\left[\frac{x\left(x-3\right)+2\left(x-3\right)}{\left(x^2+1\right)\left(x-3\right)}\right] \\ =\underset{x\to 3}{\lim }\left[\frac{\left(x+2\right)\left(x-3\right)}{\left(x^2+1\right)\left(x-3\right)}\right] \\ =\frac{3+2}{3^2+1} \\ =\frac{5}{10} \\ =\frac{1}{2} \end{gathered}$$