Question

$\underset{x\to 3}{\lim }\frac{x^4-81}{x^2-9}$

Answer 1

$$\begin{gathered} \underset{x\to 3}{\lim }\left[\frac{x^4-81}{x^2-9}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{0}{0}. \\ \underset{x\to 3}{\lim }\left[\frac{{\left(x^2\right)}^2-9^2}{x^2-9}\right] \\ =\underset{x\to 3}{\lim }\left[\frac{\left(x^2-9\right)\left(x^2+9\right)}{x^2-9}\right] \\ =\underset{x\to 3}{\lim }\left(x^2+9\right) \\ =3^2+9 \\ =18 \end{gathered}$$