Помогите пожалуйста решить 13 задание

[latex]1)\; \; \lim\limits _{n \to \infty} \frac{2n+3}{3n-4} =\Big [ \frac{:n}{:n} \Big ]= \lim\limits _{n \to \infty} \frac{2+\frac{3}{n}}{3-\frac{4}{n}}=\frac{2}{3} \\\\2)\; \; \lim\limits _{n \to \infty} \frac{n-3n^2+44}{6n^2-3n+7}=\Big [\frac{:n^2}{:n^2} \Big ] =\lim\limits _{n \to \infty} \frac{-3+\frac{1}{n}+\frac{44}{n^2}}{6-\frac{3}{n}+\frac{7}{n^2}} = \frac{-3}{6} = -\frac{1}{2} [/latex] [latex]3)\; \; \lim\limits _{n \to \infty}\frac{sin(n+3)}{n+3} =\\\\=\Big [-1 \leq sin(n+3) \leq 1\; -ogranichennaya \; ,\; n+3\to \infty \Big ]=0[/latex] [latex]4)\; \; \lim\limits _{n \to \infty} \frac{1+2+...+n}{n^2} =\\\\=\Big [1+2+...+n=\frac{1+n}{2}\cdot n=\frac{n(n+1)}{2}\; -\; arifmet.progressiya\Big ]=\\\\= \lim\limits _{n \to \infty} \frac{n(n+1)}{2n^2} = \lim\limits _{n \to \infty} \frac{n^2+n}{2n^2}=\Big [\frac{:n^2}{:n^2} \Big ]= \lim\limits _{n \to \infty} \frac{1+\frac{1}{n}}{2} =\frac{1}{2}[/latex]