Question

Вычислить пределы......................................................:

Answer 1

1)

$limlimits_{n to infty} frac{(3n - 1)(2n + 4)(n - 1)}{n^2 + n + 1} =  limlimits_{n to infty} frac{n^3(3 - frac{1}{n})(2 + frac{4}{n})(1 - frac{1}{n})}{n^2(1 + frac{1}{n} + frac{1}{n^2}} =  limlimits_{n to infty} 6n = infty$

2)

$limlimits_{n to infty} frac{sqrt{n^2 + 1} - sqrt[3]{n^2 + 1}}{sqrt[4]{n^4 + 1} + sqrt[5]{n^4 + 1}} =  limlimits_{n to infty} frac{n(sqrt{1 + frac{1}{n^2}} - sqrt[3]{frac{1}{n} + frac{1}{n^3}})}{n(sqrt[4]{1 + frac{1}{n^4}} + sqrt[5]{frac{1}{n} + frac{1}{n^5}})} =  limlimits_{n to infty} 1 = 1$

3)

$limlimits_{n to infty} (sqrt{frac{5n}{4n + 3}})^{-frac{1}{2}} = limlimits_{n to infty} sqrt[4]{frac{4n + 3}{5n}} = sqrt[4]{limlimits_{n to infty} frac{4n + 3}{5n}} = sqrt[4]{limlimits_{n to infty} frac{4}{5}} = sqrt[4]{frac{4}{5}}$

4) Тут у нас неопределённость вида либо $infty - infty$, либо $infty * 0$

$limlimits_{n to infty} (sqrt{n + 2} - sqrt{n}) = limlimits_{n to infty} frac{n + 2 - n}{sqrt{n + 2} + sqrt{n}} = limlimits_{n to infty} frac{2}{sqrt{n + 2} + sqrt{n}} = limlimits_{n to infty} frac{2}{sqrt{n}(sqrt{1 + frac{2}{n}} + 1)} = 0$

5)

$limlimits_{n to infty} (frac{1 + 2 + ldots + n}{n + 2} - frac{n}{2}) = limlimits_{n to infty} (frac{n(n + 1)}{2(n + 2)} - frac{n}{2}) = limlimits_{n to infty} frac{n^2 + n - n^2 - 2n}{2(n + 2)} = limlimits_{n to infty} -frac{n}{2(n + 2)} = -frac{1}{2}$