Question

1) Mostre, usando a definição de limites, que \underset{x\rightarrow3}{lim}\;\frac{x^2-9}{x-3}=6

Answer 1

$\underset{x\rightarrow3}{lim}\;\frac{x^2-9}{x-3}=6$

$\underset{x\rightarrow3}{lim}\;\frac{x^2-9}{x-3}$

$\underset{x\rightarrow3}{lim}\;\frac{(x+3)(x-3)}{x-3}$

$\underset{x\rightarrow3}{lim}\;x+3$

$3+3$

$6$