oblicz granicę

[alibaba8000]

1. Oblicz granicę
a) \(lim( \sqrt{n^2+5n}-n)\)
b) \(lim( \frac{3n^3+5n^2+6}{3n^2+1}- \frac{2n^3-4n+1}{2n^2-1})^ \frac{-3}{}\)
c) \(lim( \frac{1}{ \sqrt{3} }- \frac{1}{3}+ \frac{1}{3 \sqrt{3} }- \frac{1}{9}+...+ \frac{1}{3^ \frac{n-1}{} * \sqrt{3} }- \frac{1}{3^n})\)

[eresh]

1. Oblicz granicę
a) \(lim( \sqrt{n^2+5n}-n)\)

\(\Lim_{n\to\infty}a_n=\Lim_{n\to\infty}\frac{n^2+5n-n^2}{\sqrt{n^2+5n}+n}=\Lim_{n\to\infty}\frac{5}{\sqrt{1+\frac{5}{n}}+1}=\frac{5}{2}\)

[eresh]

1. Oblicz granicę

b) \(lim( \frac{3n^3+5n^2+6}{3n^2+1}- \frac{2n^3-4n+1}{2n^2-1})^ \frac{-3}{}\)

\(\Lim_{n\to\infty}( \frac{3n^3+5n^2+6}{3n^2+1}- \frac{2n^3-4n+1}{2n^2-1})^{-3}=\\
=\Lim_{n\to\infty}(\frac{6n^5-3n^3+10n^4-5n^2+12n^2-6-6n^5+12n^3-3n^2-2n^3+4n-1}{(3n^2+1)(2n^2-1)})^{-3}=\\
=\Lim_{n\to\infty}(\frac{10n^4+7n^3+4n^2+4n-7}{(3n^2+1)(2n^2-1)})^{-3}=\\
=\Lim_{n\to\infty}(\frac{10+\frac{7}{n}+\frac{4}{n^2}+\frac{4}{n^3}-\frac{7}{n^4}}{(3+\frac{1}{n^2})(2-\frac{1}{n^2})})^{-3}=(frac{10}{6})^{-3}=\frac{27}{125}\)

[eresh]

1. Oblicz granicę

c) \(lim( \frac{1}{ \sqrt{3} }- \frac{1}{3}+ \frac{1}{3 \sqrt{3} }- \frac{1}{9}+...+ \frac{1}{3^ \frac{n-1}{} * \sqrt{3} }- \frac{1}{3^n})\)

\(\Lim_{n\to\infty}(\frac{\sqrt{3}-1}{3}+\frac{\sqrt{3}-1}{9}+...+\frac{\sqrt{3}-1}{3^n})=\Lim_{n\to\infty}\frac{\frac{\sqrt{3}-1}{3}(1-(\frac{1}{3})^n)}{1-\frac{1}{3}}=\frac{\sqrt{3}-1}{3}\cdot\frac{3}{2}=\frac{\sqrt{3}-1}{2}\)

[Galen]

c)
Można też inaczej:
\(a_1= \frac{1}{ \sqrt{3} } \\q=- \frac{1}{ \sqrt{3} }\\S_n=a_1 \cdot \frac{1-q^{n-1}}{1-q}\)
\(S_n= \frac{1}{ \sqrt{3} } \cdot \frac{1-(- \frac{1}{ \sqrt{3}})^{n-1} }{1-(- \frac{1}{ \sqrt{3} }) }= \frac{1}{ \sqrt{3} } \cdot \frac{1-(- \frac{1}{ \sqrt{3}})^{n-1} }{1+ \frac{1}{ \sqrt{3} } }= \frac{1-(- \frac{1}{ \sqrt{3} } )^{n-1}}{ \sqrt{3}+1 }\)
\(\Lim_{n\to \infty }S_n= \frac{1-0}{ \sqrt{3}+1 }= \frac{1}{ \sqrt{3}+1 } \cdot \frac{ \sqrt{3}-1 }{ \sqrt{3}-1 }= \frac{ \sqrt{3}-1 }{3-1}= \frac{ \sqrt{3}-1 }{2}\)