Rozwiązanie:
[tex]d)[/tex]
[tex]$ \lim_{n \to \infty}\Big(\sqrt[3]{1-n^{3}}+n \Big) =\lim_{n \to \infty} \frac{(\sqrt[3]{1-n^{3}}+n)(\sqrt[3]{(1-n^{3})^{2}}-n\sqrt[3]{1-n^{3}} +n^{2}) }{\sqrt[3]{(1-n^{3})^{2}}-n\sqrt[3]{1-n^{3}} +n^{2}} =[/tex]
[tex]$ =\lim_{n \to \infty} \frac{1-n^{3}+n^{3}}{\sqrt[3]{(1-n^{3})^{2}}-n\sqrt[3]{1-n^{3}} +n^{2}} = \lim_{n \to \infty} \frac{1}{\sqrt[3]{(1-n^{3})^{2}}-n\sqrt[3]{1-n^{3}} +n^{2}} =[/tex]
[tex]$= \frac{1}{\infty} =0[/tex]
[tex]e)[/tex]
[tex]$ \lim_{n \to \infty} \Big(\sqrt{n+\sqrt{n} }-\sqrt{n-\sqrt{n} } \Big)=\lim_{n \to \infty} \frac{(\sqrt{n+\sqrt{n} }-\sqrt{n-\sqrt{n} } )(\sqrt{n+\sqrt{n} }+\sqrt{n-\sqrt{n} } )}{\sqrt{n+\sqrt{n} }+\sqrt{n-\sqrt{n} } } =[/tex][tex]$= \lim_{n \to \infty} \frac{n+\sqrt{n}-n+\sqrt{n} }{\sqrt{n+\sqrt{n} }+\sqrt{n-\sqrt{n} } } = \lim_{n \to \infty}\frac{2\sqrt{n} }{\sqrt{n+\sqrt{n} }+\sqrt{n-\sqrt{n} } } =[/tex]
[tex]$= \lim_{n \to \infty} \frac{2\sqrt{n} }{\sqrt{n}\Big(\sqrt{1+\frac{1}{\sqrt{n} } }\Big)+\sqrt{n}\Big(\sqrt{1-\frac{1}{\sqrt{n} } }\Big) } = \lim_{n \to \infty} \frac{2 }{\Big(\sqrt{1+\frac{1}{\sqrt{n} } }\Big)+\Big(\sqrt{1-\frac{1}{\sqrt{n} } }\Big) } =[/tex][tex]=1[/tex]
[tex]f)[/tex]
[tex]$ \lim_{n \to \infty} n\Big(2n-\sqrt{4n^{2}-3} \Big)= \lim_{n \to \infty} \frac{n\Big(2n-\sqrt{4n^{2}-3} \Big)\Big(2n+\sqrt{4n^{2}-3} \Big)}{2n+\sqrt{4n^{2}-3}} =[/tex]
[tex]$= \lim_{n \to \infty} \frac{n\Big(4n^{2}-4n^{2}+3\Big)}{2n+\sqrt{4n^{2}-3} } = \lim_{n \to \infty} \frac{3n}{n\Big(2+\sqrt{4-\frac{3}{n^{2}} }\Big) } =\frac{3}{2+2}=\frac{3}{4}[/tex]
[tex]g)[/tex]
[tex]$ \lim_{n \to \infty} n \ln\Big(\frac{n+7}{n} \Big)= \lim_{n \to \infty} \frac{\ln\Big(\frac{n+7}{n} \Big)}{\frac{1}{n} }[/tex]
Korzystamy z reguły de l'Hospitala:
[tex]$ \lim_{n \to \infty} \frac{\ln\Big(\frac{n+7}{n} \Big)}{\frac{1}{n} }= \lim_{n \to \infty} \frac{\frac{n}{n+7} \cdot \Big(-\frac{7}{n^{2}} \Big) }{-\frac{1}{n^{2}} } = \lim_{n \to \infty} \frac{7n}{n+7} =7[/tex]
[tex]h)[/tex]
[tex]$ \lim_{n \to \infty} \Big(\frac{n^{2}+1}{n^{2}} \Big)^{\Big(\left\begin{array}{c}n&2\end{array}\right \Big)}= \lim_{n \to \infty} \Big(1+\frac{1}{n^{2}} \Big)^{\frac{n^{2}-n}{2} }= \lim_{n \to \infty} [\Big(1+\frac{1}{n^{2}} \Big)^{n^{2}}]^{\frac{n^{2}-n}{2n^{2}} }=[/tex]
[tex]$= \lim_{n \to \infty} e^{\frac{n^{2}-n}{2n^{2}} }=\sqrt{e}[/tex]
[tex]g)[/tex]
[tex]$ \lim_{n \to \infty} n\Big(\frac{1}{n^{2}+1}+\frac{1}{n^{2}+2}+...+\frac{1}{n^{2}+n} \Big)= 1[/tex]
Ostatnie pomyśl, bo nie mam czasu.
