Odpowiedź :
a)
[tex]\lim_{n \to \infty} (\sqrt{3n+2}-\sqrt{3n+1})= \lim_{n \to \infty} \frac{(\sqrt{3n+2}-\sqrt{3n+1})(\sqrt{3n+2}+\sqrt{3n+1})}{\sqrt{3n+2}+\sqrt{3n+1}}=\\\\= \lim_{n \to \infty} \frac{(3n+2)-(3n+1)}{\sqrt{3n+2}+\sqrt{3n+1}}= \lim_{n \to \infty} \frac{1}{\sqrt{3n+2}+\sqrt{3n+1}}=[\frac{1}{\infty}]=0[/tex]
b)
[tex]\lim_{n \to \infty} (\sqrt{4n-3}-\sqrt{2n+10})= \lim_{n \to \infty} \frac{(\sqrt{4n-3}-\sqrt{2n+10})(\sqrt{4n-3}+\sqrt{2n+10})}{\sqrt{4n-3}+\sqrt{2n+10}}=\\\\= \lim_{n \to \infty} \frac{(4n-3)-(2n+10)}{\sqrt{4n-3}+\sqrt{2n+10}}=\lim_{n \to \infty} \frac{2n-13}{\sqrt{4n-3}+\sqrt{2n+10}}=\lim_{n \to \infty} \frac{2-\frac{13}{n}}{\sqrt{\frac{4}{n}-\frac{3}{n^2}}+\sqrt{\frac{2}{n}+\frac{10}{n^2}}}=\\=[\frac{2}{0}]=\infty[/tex]c)
[tex]\lim_{n \to \infty} (\sqrt{n^2+2n-1}-n)= \lim_{n \to \infty} \frac{(\sqrt{n^2+2n-1}-n)(\sqrt{n^2+2n-1}+n)}{\sqrt{n^2+2n-1}+n}=\\\\=\lim_{n \to \infty} \frac{(n^2+2n-1)-n^2}{\sqrt{n^2+2n-1}+n}=\lim_{n \to \infty} \frac{2n-1}{\sqrt{n^2+2n-1}+n}=\lim_{n \to \infty} \frac{2-\frac{1}{n}}{\sqrt{1+\frac{2}{n}-\frac{1}{n^2}}+1}=\\\\=\frac{2}{1+1}=1[/tex]
d)
[tex]\lim_{n \to \infty} (\sqrt{n^2+2}-\sqrt{2n^2+3})= \lim_{n \to \infty} \frac{(\sqrt{n^2+2}-\sqrt{2n^2+3})(\sqrt{n^2+2}+\sqrt{2n^2+3})}{\sqrt{n^2+2}+\sqrt{2n^2+3}}=\\\\=\lim_{n \to \infty} \frac{(n^2+2)-(2n^2+3)}{\sqrt{n^2+2}+\sqrt{2n^2+3}}=\lim_{n \to \infty} \frac{-n^2-1}{\sqrt{n^2+2}+\sqrt{2n^2+3}}=\lim_{n \to \infty} \frac{-1-\frac{1}{n^2}}{\sqrt{\frac{1}{n^2}+\frac{2}{n^4}}+\sqrt{\frac{2}{n^2}+\frac{3}{n^4}}}=\\=[\frac{-1}{0}]=-\infty[/tex]e)
[tex]\lim_{n \to \infty} (\sqrt{2n^2+1}-\sqrt{n^2+2})= \lim_{n \to \infty} \frac{(\sqrt{2n^2+1}-\sqrt{n^2+2})(\sqrt{2n^2+1}+\sqrt{n^2+2})}{\sqrt{2n^2+1}+\sqrt{n^2+2}}=\\\\=\lim_{n \to \infty} \frac{(2n^2+1)-(n^2+2)}{\sqrt{2n^2+1}+\sqrt{n^2+2}}=\lim_{n \to \infty} \frac{n^2-1}{\sqrt{2n^2+1}+\sqrt{n^2+2}}=\lim_{n \to \infty} \frac{1-\frac{1}{n^2}}{\sqrt{\frac{2}{n^2}+\frac{1}{n^4}}+\sqrt{\frac{1}{n^2}+\frac{2}{n^4}}}=\\=[\frac{1}{0}]=\infty[/tex]f)
[tex]\lim_{n \to \infty} (\sqrt{2n^2-n}-\sqrt{2n^2+3n+4})= \lim_{n \to \infty} \frac{(\sqrt{2n^2-n}-\sqrt{2n^2+3n+4})(\sqrt{2n^2-n}+\sqrt{2n^2+3n+4})}{\sqrt{2n^2-n}+\sqrt{2n^2+3n+4}}=\\=\lim_{n \to \infty} \frac{(2n^2-n)-(2n^2+3n+4)}{\sqrt{2n^2-n}+\sqrt{2n^2+3n+4}}=\lim_{n \to \infty} \frac{-4n-4}{\sqrt{2n^2-n}+\sqrt{2n^2+3n+4}}=\lim_{n \to \infty} \frac{-4-\frac{4}{n}}{\sqrt{2-\frac{1}{n}}+\sqrt{2+\frac{3}{n}+\frac{4}{n^2}}}=[/tex][tex]=\frac{-4}{\sqrt2+\sqrt2}=\frac{-4}{2\sqrt2}=\frac{-2}{\sqrt2}*\frac{\sqrt2}{\sqrt2}=\frac{-2\sqrt2}{2}=-\sqrt2[/tex]
