Question

$\underset{x\to 3}{lim}\frac{x^3-6x-9}{x^4-81}$

Answer 1

$\underset{x\to 3}{lim}\frac{x^2-6x+9}{x^4-81}$ $\frac{0}{0}$ form

Now applying L'Hospital's rule we get,

$\underset{x\to 3}{lim}\frac{2x-6}{4x^3}$

$=0$.