Question

Как такое решать хотя-бы первые 4?

Answer 1

$1)\; \; \lim\limits _{n \to \infty}\frac{\sqrt{n}}{\sqrt{n+1}}=\lim\limits _{n \to \infty}\sqrt{\frac{n}{n+1}}=\lim\limits _{n \to \infty}\sqrt{1-\frac{1}{n+1}}=\sqrt{1-0}=1$

$2)\; \; \lim\limits _{n \to \infty}\frac{2n}{\sqrt{n^2+1}+\sqrt{n^2-1}}=\Big [\; \frac{:n}{:n}\; \Big ]=\lim\limits _{n \to \infty}\frac{\frac{2n}{n}}{\sqrt{\frac{n^2+1}{n^2}}+\sqrt{\frac{n^2-1}{n^2}}}=\\\\=\lim\limits _{n \to \infty}\frac{2}{\sqrt{1+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n^2}}}=\frac{2}{\sqrt1+\sqrt1}=\frac{2}{1+1}=1$

$3)\; \; \lim\limits _{n \to \infty}\frac{\sqrt{n+2}}{\sqrt{n} +\sqrt{n+1}}=\Big [\; \frac{:\sqrt{n}}{:\sqrt{n}}\; \Big ]=\lim\limits _{n \to \infty}\frac{\sqrt{\frac{n+2}{n}}}{\sqrt{\frac{n}{n}}+\sqrt{\frac{n+1}{n}}}=\lim\limits _{n \to \infty}\frac{\sqrt{1+\frac{2}{n}}}{\sqrt1+\sqrt{1+\frac{1}{n}}}=\\\\=\frac{\sqrt1}{1+\sqrt1}=\frac{1}{2}$

$4)\; \; \lim\limits _{n \to \infty}(\sqrt{n^2+n}+\sqrt{n^2-n})=\lim\limits _{n \to \infty}\frac{(\sqrt{n^2+n}+\sqrt{n^2-n})(\sqrt{n^2+n}-\sqrt{n^2-n})}{\sqrt{n^2+n}-\sqrt{n^2-n}}=\\\\=\lim\limits _{n \to \infty}\frac{(n^2+n)-(n^2-n)}{\sqrt{n^2+n}-\sqrt{n^2-n}}=\lim\limits _{n \to \infty}\frac{2n}{\sqrt{n^2+n}-\sqrt{n^2-n}}=\Big [\; \frac{:n}{:n}\; \Big ]=$

$=\lim\limits _{n \to \infty}\frac{\frac{2n}{n}}{\sqrt{\frac{n^2+n}{n^2}}-\sqrt{\frac{n^2-n}{n^2}}}=\lim\limits _{n \to \infty}\frac{2}{\sqrt{1+\frac{1}{n}}-\sqrt{1-\frac{1}{n}}}=\Big [\; \frac{2}{1-1}=\frac{2}{0}\; \Big ]=\infty$