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Granice w nieskonczonosci

: 19 lis 2017, 22:28
autor: MartyQQ
Obliczyc:

A. \(\Lim_{n\to \infty} (\sqrt[3]{n^3+8}-\sqrt{n^2+4})\)

B. \(\Lim_{n\to \infty}(\frac{2n^2+1}{3n^2+1})^{n-n^2}\)

: 20 lis 2017, 05:37
autor: kerajs
A.
\(\Lim_{n\to \infty} (\sqrt[3]{n^3+8}-\sqrt{n^2+4})\)
\(a=\sqrt[3]{n^3+8}\\
b=\sqrt{n^2+4}\)

Możesz zastosować wzorek:
\(a^6-b^6=(a-b)(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5)\)

B.
\(\Lim_{n\to \infty}(\frac{2n^2+1}{3n^2+1})^{n-n^2}=\Lim_{n\to \infty}(\frac{2+ \frac{1}{n^2} }{3+\frac{1}{n^2} })^{-n^2(1-\frac{1}{n} )}= \left( \frac{2}{3}\right) ^{- \infty } = \infty\)