Question

Вычисление пределов. Решите пожалуйста минимум 3 примера.

Answer 1

1)
$\lim_{x \to 6} \frac{x^2-13x+42}{x^2-5x+1} =\frac{6*6-13*6+42}{6*6-5*6+1}=\frac{0}{7} = 0$

2)
$\lim_{x \to -3} \frac{x^2-9}{x^2-4x-21} = \frac{(x-3)(x+3)}{(x-7)(x+3)} = \frac{x-3}{x-7} = \frac{-3-3}{-3-7} = \frac{-6}{-10} = 0.6$

3)
$\lim_{x \to 4} \frac{x^2-8x+16}{x^2-16} = \frac{(x-4)^2}{(x-4)(x+4)}=\frac{x-4}{x+4} = \frac{4-4}{4+4} = 0$

4)
$\lim_{x \to -1} \frac{x^2+6x+5}{x^4-1} = \frac{(x+1)(x+5)}{(x^2-1)(x^2+1)} =\frac{(x+1)(x+5)}{(x-1)(x+1)(x^2+1)} = \frac{x+5}{(x-1)(x^2+1)} = \\\\ = \frac{-1+5}{(-1-1)*(1+1)} = \frac{4}{-4} = -1$