Question

Решить два предела(матанализ)

Answer 1

$1)\; \; \lim\limits _{n \to \infty}\frac{n+1}{\sqrt{4n^2+1}}=\lim\limits _{n \to \infty}\frac{1+\frac{1}{n}}{\sqrt{4+\frac{1}{n^2}}}=\frac{1+0}{\sqrt{4+0}}=\frac{1}{2}\\\\\\2)\; \; \lim\limits _{n \to \infty}(\sqrt{4n^2+n+1}-2n)=\lim\limits_{n \to \infty}\frac{(\sqrt{4x^2+n+1}-2n)(\sqrt{4x+2+n+1}+2n)}{\sqrt{4x^2+n+1}+2n}=\\\\=\lim\limits _{n \to \infty}\frac{4n^2+n+1-4n^2}{\sqrt{4n^2+n+1}+2n}=\lim\limits _{n \to \infty}\frac{n+1}{\sqrt{4n^2+n+1}+2n}=\lim\limits _{n\to \infty }\frac{1+\frac{1}{n}}{\sqrt{4+\frac{1}{n}+\frac{1}{n^2}}+2}=$

$=\frac{1+0}{\sqrt{4+0+0}+2}=\frac{1}{2+2}=\frac{1}{4}$