Question

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Answer 1

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

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