Question

calcule o limite:
$lim_{x=3} = \frac{ x^{2} -8x+15}{ x^{2}-9 }$

Answer 1

Aplique a regra de L'Hospital:

$\boxed{ \lim_{x \to 3} \frac{x^2-8x+15}{x^2-9}= \lim_{x \to 3} \frac{2x-8}{2x}=\frac{2.3-8}{2.3}=-\frac{1}{3}}$

Answer 2

$lim_{x=3} \frac{x^2-8x+15 }{x^2-9}=lim_{x=3} \frac{(x - 3)(x-5)}{(x-3)(x+3)}=lim_{x=3} \frac{x-5}{x+3}=\frac{3-5}{3+3}=\frac{-2^{:2}}{6_{:2}}=-\frac{1}{3}$