oblicz granice

[majka92]

\(\lim_{x\to \infty } \frac{2(n-1)(n+2)}{4n^2+7n-8}\)


\(\lim_{x\to \infty } \frac{n^3}{n!}\)

\(\lim_{x\to \infty } ( \sqrt{5n^2-4n+2}- \sqrt{5n^2-3}\)

\(\lim_{x\to \infty } \frac{ \sqrt[3]{ (n+1)n(n-1)} }{4n+9}\)

\(\lim_{x\to \infty } \frac{4 \circ 3^{n+1}-7 \circ 4^{n-2}}{6 \circ 2^{2n}+3^{n+8}\)

[Lbubsazob]

\(\lim_{n\to \infty } \left( \sqrt{5n^2-4n+2}- \sqrt{5n^2-3}\right) =\lim_{n\to \infty } \frac{\left( \sqrt{5n^2-4n+2}- \sqrt{5n^2-3}\right) \left( \sqrt{5n^2-4n+2}+ \sqrt{5n^2-3}\right) }{\left( \sqrt{5n^2-4n+2}+\sqrt{5n^2-3}\right) }= \\ = \lim_{n \to\infty } \frac{5n^2-4n+2-5n^2+3}{\sqrt{5n^2-4n+2}+ \sqrt{5n^2-3}}= \lim_{n \to\infty } \frac{-4n+5}{ n \sqrt{5- \frac{4}{n}+ \frac{2}{n^2} }+n \sqrt{5- \frac{3}{n^2} } }=- \frac{2}{\sqrt5}\)

[Lbubsazob]

\(\lim_{n\to \infty } \frac{2(n-1)(n+2)}{4n^2+7n-8} = \lim_{ x\to\infty } \frac{2n^2+2n-4}{4n^2+7n-8}= \lim_{n \to \infty} \frac{n^2\left( 2+ \frac{2}{n}- \frac{4}{n^2} \right) }{n^2\left( 4+ \frac{7}{n}- \frac{8}{n^2} \right) } = \frac{2}{4}=\frac{1}{2}\)

[Lbubsazob]

\(\lim_{n\to\infty }\frac{n^3}{n!}=0\)
ponieważ \(n!\) zbiega do nieskończoności szybciej niż \(n^3\).

[Lbubsazob]

\(\lim_{n\to \infty } \frac{ \sqrt[3]{ (n+1)n(n-1)} }{4n+9}=\lim_{n\to \infty } \frac{ \sqrt[3]{n^3-n} }{4n+9}= \lim_{n \to \infty } \frac{n \sqrt[3]{1- \frac{1}{n^2} } }{4n+9}=\frac{1}{4}\)

Co ma oznaczać to kółko w ostatnim przykładzie?

[majka92]

razy

[Lbubsazob]

\(\lim_{n\to \infty } \frac{4 \cdot 3^{n+1}-7 \cdot 4^{n-2}}{6 \cdot 2^{2n}+3^{n+8}}=\lim_{n\to \infty } \frac{4 \cdot 3^n \cdot \frac{1}{3}-7 \cdot 4^n \cdot 4^{-2} }{6 \cdot 4^n+3^n \cdot 3^8}= \lim_{n \to \infty} \frac{4^n\left( 4 \cdot \frac{3^n}{4^n} \cdot \frac{1}{3}-7 \cdot 4^{-2} \right) }{4^n\left( 6+ \frac{3^n}{4^n} \cdot 3^8 \right) }= \frac{-7 \cdot \frac{1}{16} }{6}=- \frac{7}{96}\)

BTW, symbol mnożenia to \cdot.