[latex] a) lim_{n o infty} (sqrt{3n+2}-sqrt{3n+1})= \ = lim_{n o infty} (sqrt{3n+2}-sqrt{3n+1}) cdot frac{sqrt{3n+2}+sqrt{3n+1}}{sqrt{3n+2}+sqrt{3n+1}} = \ = lim_{n o infty} frac{(sqrt{3n+2}-sqrt{3n+1})(sqrt{3n+2}+sqrt{3n+1})}{sqrt{3n+2}+sqrt{3n+1}} =\= lim_{n o infty} frac{(3n+2)-(3n+1)}{sqrt{3n+2}+sqrt{3n+1}}= \ = lim_{n o infty} frac{3n+2-3n-1}{sqrt{3n+2}+sqrt{3n+1}}= \ = lim_{n o infty} frac1{sqrt{3n+2}+sqrt{3n+1}}=frac1{sqrt{infty}+sqrt{ infty}}[/latex] [latex]=frac1{infty}=0[/latex] b) [latex]...= lim_{n o infty} (sqrt{n^2+2n-1}-n)= \ =lim_{n o infty} (sqrt{n^2+2n-1}-n) cdot frac{sqrt{n^2+2n-1}+n}{sqrt{n^2+2n-1}+n}= \ = lim_{n o infty} frac{n^2+2n-1-n^2}{sqrt{n^2+2n-1}+n}= lim_{n o infty} frac{2n-1}{sqrt{n^2(1+frac2{n}-frac1{n^2})}+n}= \ = lim_{n o infty} frac{2n-1}{n cdot sqrt{1+frac2{n}-frac1{n^2})}+n}= lim_{n o infty} frac{n(2-frac1n)}{n(sqrt{1+frac2n-frac1{n^2}}+1)}=\= lim_{n o infty} frac{2-frac1n}{sqrt{1+frac2n-frac1{n^2}}+1}=[/latex] [latex]=frac{2-0}{sqrt{1+0-0}+1}=frac2{1+1}=1[/latex] c) Najpierw policzmy pomocniczo granice [latex]lim_{n o infty} sqrt[n]{4^n}[/latex] oraz [latex] lim_{n o infty} sqrt[n]{4^n+4^n+4^n}[/latex] : [latex] lim_{n o infty} sqrt[n]{4^n}= lim_{n o infty} 4^{frac{n}{n}}=4^1=4 \ \ lim_{n o infty} sqrt[n]{4^n+4^n+4^n}= lim_{n o infty} sqrt[n]{3cdot 4^n}= lim_{n o infty} sqrt[n]3cdot sqrt[n]{4^n}=\= lim_{n o infty} 3^{frac1n}cdot 4^{frac{n}{n}}=3^{ lim_{n o infty} frac1n}cdot 4^1=3^0cdot 4=1cdot 4=4[/latex] Ponieważ [latex]sqrt[n]{4^n}le sqrt{2^n+3^n+4^n}le sqrt[n]{4^n+4^n+4^n}[/latex], oraz granicą ciągów [latex]sqrt[n]{4^n}[/latex] oraz [latex]sqrt[n]{4^n+4^n+4^n}[/latex] jest 4, to z twierdzenia o trzech ciągach wynika, że granicą ciągu [latex]sqrt[n]{2^n+3^n+4^n}[/latex] też jest 4.
