Question

Помогите, пожалуйста, Вычислить пределы

Answer 1

Ответ:

Пошаговое объяснение:

1)

$\displaystyle \lim_{n \to \infty} \frac{(1+n)&2}{\sqrt[3]{n^6+1} } = \lim_{n \to \infty}\sqrt[3]{\frac{(1+n)^6}{n^6+1} } =\sqrt[3]{ \lim_{n \to \infty} \frac{(1+n)^6}{n^6+1} } =$

$=\displaystyle \sqrt[3]{ \lim_{n \to \infty} \frac{n^6+6n^5+15n^4+2n^3+15n^2+6n+1}{n^6+1} } =$

$=\displaystyle \sqrt[3]{ \lim_{n \to \infty} \frac{\frac{n^6}{n^6} +6\frac{n^5}{n^6} +15\frac{n^4}{n^6} +2\frac{n^3}{n^6} +15\frac{n^2}{n^6} +6\frac{n}{n^6} +\frac{1}{n^6} }{\frac{n^6}{n^6} +\frac{1}{n^6} } } =\sqrt[3]{\frac{1}{1} } =1$

2)

$\displaystyle \lim_{x \to 0} \frac{\sqrt{5+x} -\sqrt{5} }{x} \lim_{x \to 0} \frac{(\sqrt{5+x} -\sqrt{5} ) ( \sqrt{5+x} +\sqrt{5} )}{x( \sqrt{5+x} +\sqrt{5} )} =$

$\displaystyle = \lim_{x \to 0} \frac{x}{x(\sqrt{5+x} +\sqrt{5} } = \lim_{x \to 0} \frac{1}{\sqrt{5+x} +\sqrt{5} } =\frac{1}{2\sqrt{5} } =\frac{\sqrt{5} }{10}$

3)

$\displaystyle \lim_{x \to \infty} (\sqrt{x^2+1} -\sqrt{x^2-1} ) = \lim_{x \to \infty} \frac{ (\sqrt{x^2+1} -\sqrt{x^2-1} )* (\sqrt{x^2+1} +\sqrt{x^2-1} )}{ (\sqrt{x^2+1} +\sqrt{x^2-1} )} =$

$\displaystyle = \lim_{x \to 0} \frac{2}{x\bigg (\sqrt{\frac{x^2+1}{x^2} } +\sqrt{\frac{x^2-1}{x^2} }\bigg ) } = \lim_{x \to 0} \frac{2}{x \bigg(\sqrt{1+\frac{1}{x^2} }+ \sqrt{1-\frac{1}{x^2} }\bigg )} =\frac{2}{\infty *2} =0$