Rozwiązanie:
[tex]\bold{1.}[/tex]
[tex]$ \lim_{n \to \infty} \frac{(1-2n^{2})^{2}}{1-4n^{3}+3n^{4}} = \lim_{n \to \infty} \frac{n^{4}\Big(\frac{1}{n^{2}}-2\Big)^{2}}{n^{4}\Big(\frac{1}{n^{4}}-\frac{4}{n}+3 \Big)} =\frac{4}{3}[/tex]
[tex]\bold{2.}[/tex]
[tex]$ \lim_{n \to \infty} \Big(\sqrt{n^{2}+n}-\sqrt{n^{2}-n}\Big)= \lim_{n \to \infty}\frac{\Big(\sqrt{n^{2}+n}-\sqrt{n^{2}-n}\Big)\Big(\sqrt{n^{2}+n}+\sqrt{n^{2}-n}\Big)}{\sqrt{n^{2}+n}+\sqrt{n^{2}-n}} =[/tex]
[tex]$= \lim_{n \to \infty} \frac{n^{2}+n-(n^{2}-n)}{\sqrt{n^{2}+n}+\sqrt{n^{2}-n}} = \lim_{n \to \infty} \frac{2n}{\sqrt{n^{2}+n}+\sqrt{n^{2}-n}} =[/tex]
[tex]$= \lim_{n \to \infty}\frac{2n}{n\sqrt{1+\frac{1}{n} }+n\sqrt{1-\frac{1}{n} } } = \lim_{n \to \infty} \frac{2}{\sqrt{1+\frac{1}{n} }+\sqrt{1-\frac{1}{n} } } =\frac{2}{1+1}=1[/tex]
[tex]\bold{3.}[/tex]
[tex]$ \lim_{n \to \infty} \Big(\frac{n+4}{n+1} \Big)^{n}=\lim_{n \to \infty} \Big(\frac{n+1+3}{n+1} \Big)^{n}=\lim_{n \to \infty} \Big(1+\frac{3}{n+1} \Big)^{n}=[/tex]
[tex]$=\lim_{n \to \infty} \Big[\Big(1+\frac{3}{n+1} \Big)^{\frac{n+1}{3} }\Big]^{\frac{3n}{n+1} }=\lim_{n \to \infty} e^{\frac{3n}{n+1} }=e^{3}[/tex]