1. limx^2-9:x-3 x>3 2. lim x-3: корень х+1 и -2 х>3 3. lim x^2=x-6:x-2 x>2 помогитее

Решите задачу:

$\lim_{x \to \3} \frac{ x^{2} -9}{x-3} = \frac{ 3^{2}-9 }{3-3} =[ \frac{0}{0} ]$
$\lim_{x \to 3} \frac{ x^{2} -9}{x-3} = \lim_{x \to 3} \frac{(x-3)*(x+3)}{x-3} = \lim_{x \to 3} (x+3)=3+3=6$

$\lim_{x \to 3} \frac{x-3}{ \sqrt{x+1}-2 } =[ \frac{0}{0}]$
$\lim_{x \to 3} \frac{(x-3)*( \sqrt{x+1}-2)}{( \sqrt{x+1}-2 )*( \sqrt{x+1}+2 )} = \lim_{x \to 3} \frac{(x-3)*( \sqrt{x+1}-2 )}{( \sqrt{x+1} ) ^{2}-2^{2} } =$
$= \lim_{x \to 3} \frac{(x-3)*( \sqrt{x+1}+2) }{x-3} = \lim_{x \to 3} ( \sqrt{x+1} +2)= \sqrt{3+1}+2 =4$