Question

помогите с алгеброй ​

Answer 1

Объяснение:

$\lim_{x \to 1} \frac{x^3+3x-1}{\sqrt{x^5+3} } =\frac{1^3+3*1-1}{\sqrt{1^5+3} } =\frac{1+3-1}{\sqrt{1+3} }=\frac{3}{\sqrt{4} } =\frac{3}{2}.$

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