Question

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Answer 1

$1); limlimits _{x to -2}frac{x^2-2x-8}{x^3+8}=limlimits_{x to -2}frac{(x+2)(x-4)}{(x+2)(x^2-2x+4)}=limlimits_{x to -2}frac{x-4}{x^2-2x+4}=\=frac{-2-4}{4+4+4}= frac{-6}{12}=-frac{1}{2}\\2); limlimits_{x to infty}Big (frac{7x-4}{7x+1}Big )^{2x-5}=limlimits_{x to infty}Big (Big (1+frac{-5}{7x+1}Big )^{frac{7x+1}{-5}}Big )^{frac{-5(2x-5)}{7x+1}}=\\=e^{limlimits_{x to infty}frac{-10x+25}{7x+1}}=e^{-frac{10}{7}}$

$3); limlimits_{x to 1} frac{2-sqrt{x}}{sqrt{6x+1}-5}=frac{2-sqrt1}{sqrt{7}-5}=frac{1}{sqrt7-5}\\4); limlimits_{x to infty }frac{8x^2-9x+5}{4x^2-6x+7}= limlimits_{x to infty}frac{8x^2}{4x^2}=frac{8}{4}=2\\5); limlimits_{x to infty}(sqrt{x^2-5x+7}-x)=limlimits_{x to infty}frac{(sqrt{x^2-5x+7}-x)(sqrt{x^2-5x+7}+x)}{sqrt{x^2-5x+7}+x}=\\=limlimits_{x to infty}frac{x^2-5x+7-x^2}{sqrt{x^2-5x+7}+x}=limlimits _{x to infty}frac{7-5x}{sqrt{x^2-5x+7}+x}}=$

$= limlimits_{x to infty}frac{frac{7}{x}-5}{sqrt{1-frac{5}{x}+frac{7}{x^2}}+1}=frac{-5}{1+1}=-2,5\\6); limlimits _{x to 0}frac{arcsin7x}{5x}=[, arcsin alpha (x)sim alpha (x),; alpha (x)to 0, ]=\\=limlimits _{x to 0}, frac{7x}{5x}=frac{7}{5}$