Question

помогите,пожалуйста)!

Answer 1

$1); limlimits _{xto 0}, frac{sin6x-sin8x}{x} =limlimits _{xto 0} frac{-2sinxcdot cos7x}{x} =\\=-2limlimits _{xto 0}(frac{sinx}{x}cdot cos7x)=-2cdot 1cdot 1=-2 \\2) ; limlimits _{x to infty}, (1+frac{1}{2x^2})^{2x^2}=e; ; quad [; ; limlimits_{alpha (x) to infty} (1+frac{1}{alpha (x)})^{alpha (x)}=e; ; ; ]$

$3); ; lim_{x to 0} frac{1-cosx}{x^2}=limlimits _{xto 0}frac{2sin^2frac{x}{2}}{x^2} =2limlimits _{xto 0}frac{sinfrac{x}{2}}{frac{x}{2} }cdot frac{sinfrac{x}{2}}{frac{x}{2}}cdot frac{1}{4}=frac{1}{4}$

$4); ; limlimits _{x to infty} (frac{2x+1}{2x-3} )^{2x+1}= limlimits _{x to infty} (1+frac{4}{2x-3} )^{frac{2x-3}{4}cdot frac{4(2x+1)}{2x-3} }=\\=e^{limlimits _{xto infty }frac{8x+4}{2x-3}}=e^{frac{8}{2}}=e^4$