Pamiętajmy, że:

$\infty -\infty$  

to symbol nieoznaczony, zatem w każdym przykładzie musimy przekształcić wyrażenie tak aby się pozbyć powyższego symbolu.

 

$a)a_n=\sqrt{n}-\sqrt{n+1}=\left(\sqrt{n}-\sqrt{n+1}\right)\cdot \frac{\sqrt{n}+\sqrt{n+1}}{\sqrt{n}+\sqrt{n+1}}=\frac{{\left(\sqrt{n}\right)}^2-{\left(\sqrt{n+1}\right)}^2}{\sqrt{n}+\sqrt{n+1}}=\frac{n-\left(n+1\right)}{\sqrt{n}+\sqrt{n+1}}=-\frac{1}{\sqrt{n}+\sqrt{n+1}}\stackrel{\left[-\frac{1}{\infty }\right]}{\underset{n\to \infty }{\to }}0$  

$b)a_n=\sqrt{2n+4}-\sqrt{2n}=\left(\sqrt{2n+4}-\sqrt{2n}\right)\cdot \frac{\sqrt{2n+4}+\sqrt{2n}}{\sqrt{2n+4}+\sqrt{2n}}=\frac{{\left(\sqrt{2n+4}\right)}^2-{\left(\sqrt{2n}\right)}^2}{\sqrt{2n+4}+\sqrt{2n}}=\frac{2n+4-2n}{\sqrt{2n+4}+\sqrt{2n}}=\frac{4}{\sqrt{2n+4}+\sqrt{2n}}\stackrel{\left[\frac{4}{\infty }\right]}{\underset{n\to \infty }{\to }}0$  

 

$c)a_n=\sqrt{n^2+n}-n=\left(\sqrt{n^2+n}-n\right)\cdot \frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}=\frac{{\left(\sqrt{n^2+n}\right)}^2-n^2}{\sqrt{n^2+n}+n}=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2\left(1+\frac{1}{n}\right)}+n}=\frac{n}{n\cdot \left(\sqrt{1+\frac{1}{n}}+1\right)}=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\underset{n\to \infty }{\to }\frac{1}{1+1}=\frac{1}{2}$  

 

$d)a_n=\sqrt{3n^2-2n}-\sqrt{3n^2-1}=\left(\sqrt{3n^2-2n}-\sqrt{3n^2-1}\right)\cdot \frac{\sqrt{3n^2-2n}+\sqrt{3n^2-1}}{\sqrt{3n^2-2n}+\sqrt{3n^2-1}}=\frac{{\left(\sqrt{3n^2-2n}\right)}^2-{\left(\sqrt{3n^2-1}\right)}^2}{\sqrt{3n^2-2n}+\sqrt{3n^2-1}}=\frac{3n^2-2n-3n^2+1}{\sqrt{3n^2-2n}+\sqrt{3n^2-1}}=\frac{-2n+1}{\sqrt{3n^2-2n}+\sqrt{3n^2-1}}=\frac{n\left(-2+\frac{1}{n}\right)}{\sqrt{n^2\left(3-\frac{2}{n}\right)}+\sqrt{n^2\left(3-\frac{1}{n}\right)}}=\frac{n\left(-2+\frac{1}{n}\right)}{n\sqrt{3-\frac{2}{n}}+n\sqrt{3-\frac{1}{n}}}=\frac{n\left(-2+\frac{1}{n}\right)}{n\left(\sqrt{3-\frac{2}{n}}+\sqrt{3-\frac{1}{n}}\right)}=\frac{-2+\frac{1}{n}}{\sqrt{3-\frac{2}{n}}+\sqrt{3-\frac{1}{n}}}\stackrel{\left[\frac{\sqrt{5}}{\infty }\right]}{\underset{n\to \infty }{\to }}\frac{-2+0}{\sqrt{3-0}+\sqrt{3-0}}=-\frac{2}{\sqrt{3}+\sqrt{3}}=-\frac{2}{2\sqrt{3}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$  

 

$e)a_n=\sqrt{n+5}-\sqrt{n}=\left(\sqrt{n+5}-\sqrt{n}\right)\cdot \frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+5}+\sqrt{n}}=\frac{{\left(\sqrt{n+5}\right)}^2-{\left(\sqrt{n}\right)}^2}{\sqrt{n+5}+\sqrt{n}}=\frac{n+5-n}{\sqrt{n+5}+\sqrt{n}}=\frac{5}{\sqrt{n+5}+\sqrt{n}}\stackrel{\left[\frac{5}{\infty }\right]}{\underset{n\to \infty }{\to }}0$  

 

$f)a_n=\sqrt{n^2-2n}-n=\left(\sqrt{n^2-2n}-n\right)\cdot \frac{\sqrt{n^2-2n}+n}{\sqrt{n^2-2n}+n}=\frac{{\left(\sqrt{n^2-2n}\right)}^2-n^2}{\sqrt{n^2-2n}+n}=\frac{n^2-2n-n^2}{\sqrt{n^2-2n}+n}=\frac{-2n}{\sqrt{n^2-2n}+n}=\frac{n\cdot \left(-2\right)}{n\sqrt{1-\frac{2}{n}}+n}=\frac{n\cdot \left(-2\right)}{n\left(\sqrt{1-\frac{2}{n}}+1\right)}=-\frac{2}{\sqrt{1-\frac{2}{n}}+1}\left(\underset{n\to \infty }{\to }\right)-\frac{2}{\sqrt{1-0}+1}=-\frac{2}{1+1}=-\frac{2}{2}=-1$  

 

$g)a_n=\sqrt{2n^2+10n-3}-\sqrt{2n^2-2n+3}=\left(\sqrt{2n^2+10n-3}-\sqrt{2n^2-2n+3}\right)\cdot \frac{\sqrt{2n^2+10n-3}+\sqrt{2n^2-2n+3}}{\sqrt{2n^2+10n-3}+\sqrt{2n^2-2n+3}}=\frac{{\left(\sqrt{2n^2+10n-3}\right)}^2-{\left(2n^2-2n+3\right)}^2}{\sqrt{2n^2+10n-3}+\sqrt{2n^2-2n+3}}=\frac{2n^2+10n-3-2n^2+2n-3}{\sqrt{2n^2+10n-3}+\sqrt{2n^2-2n+3}}=\frac{12n-6}{\sqrt{2n^2+10n-3}+\sqrt{2n^2-2n+3}}=\frac{n\left(12-\frac{6}{n}\right)}{n\sqrt{2+\frac{10}{n}-\frac{3}{n^2}}+n\sqrt{2-\frac{2}{n}+\frac{3}{n^2}}}=\frac{n\left(12-\frac{6}{n}\right)}{n\left(\sqrt{2+\frac{10}{n}-\frac{3}{n^2}}+\sqrt{2-\frac{2}{n}+\frac{3}{n^2}}\right)}=\frac{12-\frac{6}{n}}{\sqrt{2+\frac{10}{n}-\frac{3}{n^2}}+\sqrt{2-\frac{2}{n}+\frac{3}{n^2}}}\underset{n\to \infty }{\to }\frac{12-0}{\sqrt{2+0-0}+\sqrt{2-0+0}}=\frac{12}{2\sqrt{2}}=\frac{6}{\sqrt{2}}=\frac{6\sqrt{2}}{2}=3\sqrt{2}$  

 

$h)a_n=\frac{\sqrt{n\left(n+2\right)}-n}{n+2-\sqrt{n\left(n+2\right)}}\cdot \frac{\sqrt{n\left(n+2\right)}+n}{\sqrt{n\left(n+2\right)}+n}=\frac{{\left(\sqrt{n\left(n+2\right)}\right)}^2-n^2}{\left(n+2\right)\sqrt{n\left(n+2\right)}+n\left(n+2\right)-{\left(\sqrt{n\left(n+2\right)}\right)}^2-n\sqrt{n\left(n+2\right)}}=\frac{n\left(n+2\right)-n^2}{\cancel{n\sqrt{n\left(n+2\right)}}+2\sqrt{n\left(n+2\right)}+\cancel{n\left(n+2\right)}-\cancel{n\left(n+2\right)}-\cancel{n\sqrt{n\left(n+2\right)}}}=\frac{\cancel{n^2}+2n-\cancel{n^2}}{2\sqrt{n\left(n+2\right)}}=\frac{2n}{2\sqrt{n\left(n+2\right)}}=\frac{n}{\sqrt{n^2+2n}}=\frac{n}{n\sqrt{1+\frac{2}{n}}}=\frac{1}{\sqrt{1+\frac{2}{n}}}\underset{n\to \infty }{\to }1$