Question

Помогите пожалуйста решить ?((

Answer 1

• $\lim_{x \to -2} \frac{2x^2+3x-2}{3x^2+2x-8} = \lim_{x \to -2} \frac{(2x - 1)(x + 2)}{(3 x - 4) (x + 2)} = \lim_{x \to -2} \frac{(2x - 1)}{(3 x - 4)} = \frac{-5}{-10} = \frac{1}{2}$
• $\lim_{x \to 2} \frac{\sqrt{3x-2}-2}{x^2-4} = \lim_{x \to 2} \frac{3x-6}{(x-2)(x+2)(\sqrt{3x-2} + 2)} = \lim_{x \to 2} \frac{3(x-2)}{(x-2)(x+2)(\sqrt{3x-2} + 2)} = \lim_{x \to 2} \frac{3}{(x+2)(\sqrt{3x-2} + 2)} = \frac{3}{4*4}= \frac{3}{16}$
• $\lim_{a \to \infty} \frac{2a^3-a+1}{a^2+2a-5} = \lim_{a \to \infty} \frac{2-\frac{1}{a^2} +\frac{1}{a^3}}{\frac{1}{a} +\frac{2}{a^2}-\frac{5}{a^3} } = (\frac{2}{0}) = \infty$