Question

Помогите пожалуйстаа по матем, срочно

Answer 1

$limlimits_{x to 2} frac{x^3-8}{x^2+x-6} = limlimits_{x to 2} frac{(x-2)(x^2+x+4)}{(x+3)(x-2)} = limlimits_{x to 2} frac{x^2+x+4}{x+3} =frac{2^2+2+4}{2+3} = 2;$

$limlimits_{x to 0} frac{5x}{tg4x} = limlimits_{x to 0} frac{5}{4 frac{tg4x}{4x} } = frac{5}{4} = 1,25;$

$limlimits_{x to infty} (frac{x-2}{x+1})^{2x-3} =limlimits_{x to infty} (1-frac{3}{x+1})^{2x-3} =limlimits_{x to infty} (1+frac{1}{ -frac{x+1}{3} })^{2x-3} =\= limlimits_{x to infty} ((1+frac{1}{ -frac{x+1}{3} })^{- frac{x+1}{3}})^{- frac{3}{x+1}cdot(2x-3)} = \ =limlimits_{x to infty} ((1+frac{1}{ -frac{x+1}{3} })^{- frac{x+1}{3}})^{- 3cdotfrac{2- frac{3}{x} }{1+ frac{1}{x}}} = e^{-3cdot2}=e^{-6};$

$limlimits_{x to infty} frac{3x^2+5x-7}{3x^2+x+1} = limlimits_{x to infty} frac{3+ frac{5}{x} - frac{7}{x^2}}{3+ frac{1}{x} + frac{1}{x^2}} = frac{3+0 - 0}{3+0 +0} = 1.$