Granice ciągów (3)

[lubiepaczki]

Obliczyć granice ciągów:
\(k) \lim_{n\to \infty} \frac{ \left( n+2 \right)! + \left( n+1\right)! }{ \left(n+3 \right)! }
l)\lim_{n\to \infty} \frac{ \sqrt{n+2} - \sqrt{n+1} }{ \sqrt{n+1}- \sqrt{n} }
m)\lim_{n\to \infty} n(2n- \sqrt{4n^{2}-3} )
n)\lim_{n\to \infty} \frac{(n^{2}+1)n! +1}{ \left(2n+1 \right) \left(n+1 \right)! }\)

[radagast]

\(k) \lim_{n\to \infty} \frac{ \left( n+2 \right)! + \left( n+1\right)! }{ \left(n+3 \right)! }=\lim_{n\to \infty} \frac{ \left( n+2 \right) + 1 }{ \left(n+3 \right) \left(n+2 \right) }=0\)

[radagast]

\(l)\lim_{n\to \infty} \frac{ \sqrt{n+2} - \sqrt{n+1} }{ \sqrt{n+1}- \sqrt{n} }=\lim_{n\to \infty} \frac{ (\sqrt{n+2} - \sqrt{n+1})(\sqrt{n+2} + \sqrt{n+1}) (\sqrt{n+1}+\sqrt{n} )}{ (\sqrt{n+1}- \sqrt{n} )(\sqrt{n+1}+\sqrt{n} )(\sqrt{n+2} + \sqrt{n+1}) } =
\lim_{n\to \infty} \frac{ (n+2-n-1) (\sqrt{n+1}+\sqrt{n} )}{ (n+1-n)(\sqrt{n+2} + \sqrt{n+1}) }=\lim_{n\to \infty} \frac{ (\sqrt{n+1}+\sqrt{n} )}{ (\sqrt{n+2} + \sqrt{n+1}) }=\lim_{n\to \infty} \frac{ (\sqrt{1+ \frac{1}{n} }+\sqrt{1} )}{ (\sqrt{1+ \frac{2}{n} } + \sqrt{1+ \frac{1}{n} }) }= \frac{1+1}{1+1}=1\)

[radagast]

\(m)
\lim_{n\to \infty} n(2n- \sqrt{4n^{2}-3} )=\lim_{n\to \infty} n(2n- \sqrt{4n^{2}-3} ) \cdot \frac{(2n+ \sqrt{4n^{2}-3} )}{(2n+ \sqrt{4n^{2}-3} )} =\lim_{n\to \infty} n(4n^2- 4n^{2}+3 ) \cdot \frac{1}{(2n+ \sqrt{4n^{2}-3} )} =
\lim_{n\to \infty} \frac{3n}{(2n+ \sqrt{4n^{2}-3} )} =\lim_{n\to \infty} \frac{3}{(2+ \sqrt{4- \frac{3}{n^2} } )} = \frac{3}{4}\)

[radagast]

\(n)
\lim_{n\to \infty} \frac{(n^{2}+1)n! +1}{ \left(2n+1 \right) \left(n+1 \right)! }=\lim_{n\to \infty} \frac{(n^{2}+1)+ \frac{1} {n!}}{ \left(2n+1 \right) \left(n+1 \right) }= \frac{1}{2}\)