Zadanie 1 lim(1+2/3n)^4n-8= zadanie 2 lim(√4n^2+5n-7

Peashooter 1.
tw: jeżeli an dąży do 0, bn dąży do ∞, an*bn dąży do g przy n->∞, to:
$\lim_{n \to \infty} (1+a_n)^{b_n}=e^g$
zatem
$a_n= \frac{2}{3n}, \lim_{n \to \infty} a_n =0 \\ b_n=4n-8, \lim_{n \to \infty} b_n=\infty \\ a_n*b_n= \frac{2(4n-8)}{3n} , \lim_{n \to \infty} a_n *b_n= \frac{8}{3} \\ \\ \lim_{n \to \infty} (1+ \frac{2}{3n} )^{4n-8}=e^ \frac{8}{3}$

2.
$\frac{( \sqrt{4n^2+5n-7}-2n)(\sqrt{4n^2+5n-7}+2n) }{\sqrt{4n^2+5n-7}+2n} = \frac{4n^2+5n-7-4n^2}{\sqrt{4n^2+5n-7}+2n} = \frac{5n-7}{\sqrt{4n^2+5n-7}+2n} \\ \\ = \frac{5- \frac{7}{n} }{ \sqrt{4+ \frac{5}{n} - \frac{7}{n^2} }+2 } \\ \\ \lim_{n \to \infty} a_n = \frac{5-0}{ \sqrt{4+0-0}+2 } = \frac{5}{4}$

2 votes Thanks 1

Konrad9001 1)

$\lim_{n \to \infty} (1+ \frac{2}{3n})^{4n-8}= \lim_{n \to \infty} ((1+ \frac{2}{3n})^{3n})^{ \frac{4n-8}{3n}}=(e^{2})^\frac{4}{3}=e^{\frac{8}{3}}$

$\lim_{n \to \infty} \frac{4n-8}{3n} = \frac{n(4- \frac{8}{n}) }{3n} = \frac{4}{3}$

2)

$\lim_{n \to \infty} \sqrt{4n^2+5n-7} -2n * \frac{\sqrt{4n^2+5n-7} +2n}{\sqrt{4n^2+5n-7} +2n} = \\ \frac{4n^2+5n-7-4n^2}{ \sqrt{4n^2+5n-7} +2n} = \frac{5n-7}{ \sqrt{n^2(4+ \frac{5}{n} - \frac{7}{n^2})} +2n}$
$= \frac{5n-7}{ \sqrt{n^2(4+ \frac{5}{n} - \frac{7}{n^2})} +2n} = \frac{5n-7}{ \sqrt{4n^2} + 2n} = \frac{5n-7}{4n} = \frac{n(5- \frac{7}{n})}{4n} = \frac{5}{4}$

2 votes Thanks 1

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