Question

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Answer 1

$\lim\limits _{n \to \infty}\frac{2^{n-1}-4n}{5n+2^{n-3}}=\Big [\; \frac{\infty }{\infty }\; \to \; Lopital\; \; \Big ]=\lim\limits_{n \to \infty}\frac{2^{n-1}\cdot ln2-4}{5+2^{n-3}\cdot ln2}=[\; Lopital\; ]=\\\\=\lim\limits_{n \to \infty}\frac{2^{n-1}\cdot ln^22}{2^{n-3}\cdot ln^22}=\lim\limits_{n \to \infty}\frac{2^{n}\cdot 2^{-1}}{2^{n}\cdot 2^{-3}}=\frac{2^{-1}}{2^{-3}}=2^{-1+3}=2^2=4\\\\ili$

$\lim\limits_{n \to \infty}\frac{2^{n-1}-4n}{5n+2^{n-3}}=\lim\limits_{n \to \infty}\frac{2^{n}\cdot (2^{-1}-\frac{4n}{2^{n}})}{2^{n}\cdot (\frac{5n}{2^{n}}+2^{-3})}=\lim\limits _{n \to \infty}\frac{2^{-1}\, -0}{0\, +2^{-3}}=\frac{2^{-1}}{2^{-3}}=2^2=4\\\\.\Big [\; \frac{4n}{2^{n}}\to 0\; \; ,\; \; \frac{5n}{2^{n}}\to 0\; \; pri\; \; n\to \infty \;\Big ]$