Question

ПОМОГИТЕ решить 1 вариант

Answer 1

Ответ:

1)

$\lim_{x \to -1} \frac{x^{3}+1 }{x^{2}-1}= \lim_{x \to -1} \frac{(x+1)(x^{2}-x+1) }{(x+1)(x-1)}=\\= \lim_{x \to -1} \frac{x^{2}-x+1}{x-1}=\frac{(-1)^{2}-(-1)+1}{-1-1} =\frac{1+1+1}{-2} =-\frac{3}{2}=-1,5$

2)

$\lim_{x \to 3} \frac{x^{2}-2x-3 }{x^{2}-9}= \lim_{x \to 3} \frac{(x-3)(x+1) }{(x-3)(x+3)}=\lim_{x \to 3} \frac{x+1}{x+3}=\frac{3+1}{3+3}=\frac{4}{6}=\frac{2}{3}$

3)

$\lim_{x \to \infty} \frac{4x^{2} }{x^{2} -1} = \lim_{x \to \infty} \frac{\frac{4x^{2} }{x^{2} } }{\frac{x^{2} }{x^{2} }  -\frac{1}{x^{2} } } =\lim_{x \to \infty} \frac{4}{1  -\frac{1}{x^{2} } } =\frac{4}{1  -0} =4$

4)

$\lim_{x \to \infty} \frac{x^{2}-1}{x^{2} +x} =\lim_{x \to \infty} \frac{\frac{x^{2}}{x^{2}} -\frac{1}{x^{2}}  }{\frac{x^{2}}{x^{2}} +\frac{x}{x^{2}} } =\lim_{x \to \infty} \frac{1-\frac{1}{x^{2}} }{1 +\frac{1}{x} } =\frac{1-0}{1 +0 } =1$