a) \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt{n^2+n+1}-n\right)=\)\(\lim\limits_{n\rightarrow+\infty}\dfrac{n+1}{\left(\sqrt{n^2+n+1}+n\right)}=\)\(\lim\limits_{n\rightarrow+\infty}\dfrac{n\left(1+\dfrac{1}{n}\right)}{n\left(\sqrt{1+\dfrac{1}{n}+\dfrac{1}{n^2}}+1\right)}=\)\(\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}+\dfrac{1}{n^2}}+1}=\dfrac{1}{2}\)
b) \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt{n^3+n^2+1}-\sqrt{n^2+n}\right)=\)\(\lim\limits_{n\rightarrow+\infty}\dfrac{n^3-n+1}{\left(\sqrt{n^3+n^2+1}+\sqrt{n^2+n}\right)}=\)\(\lim\limits_{n\rightarrow+\infty}\dfrac{n^3\left(1-\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)}{n^{\dfrac{3}{2}}\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n\sqrt{1+\dfrac{1}{n^2}}}=+\infty\)
c) \(\lim\limits_{n\rightarrow+\infty}\dfrac{n\left(n+1\right).sin\left(n\right)}{n^3-2n+3}=\)\(\lim\limits_{n\rightarrow+\infty}\dfrac{n^2\left(1+\dfrac{1}{n}\right).sin\left(n\right)}{n^3\left(1-\dfrac{2}{n^2}+\dfrac{3}{n^3}\right)}=\)\(\lim\limits_{n\rightarrow+\infty}\dfrac{\left(1+\dfrac{1}{n}\right).sin\left(n\right)}{n\left(1-\dfrac{2}{n^2}+\dfrac{3}{n^3}\right)}=0.sin\left(n\right)=0\)
