Question

найти предел функций, не пользуясь правилом Лопиталя
Даю 45 балов

Answer 1

а

$limlimits_{x to infty} frac{x^3 + 1}{2x^3 + 1} =limlimits_{x to infty} frac{x^3(1 + 0)}{x^3(2 + 0)} = frac{1}{2}$

б

$limlimits_{x to 7} frac{sqrt{2 + x} - 3}{x - 7}, t = x - 7, x = t + 7\limlimits_{t to 0} frac{sqrt{9 + t} - 3}{t} = limlimits_{t to 0} frac{3(sqrt{1 + frac{t}{9}} - 1)}{t} = limlimits_{t to 0} frac{3frac{t}{18}}{t} = frac{1}{6}$

в

$limlimits_{x to 0} frac{sin(3x)}{5x} = limlimits_{x to 0} frac{3x sin(3x)}{5x cdot 3x} = limlimits_{x to 0} frac{3x}{5x} = frac{3}{5}$

г

$limlimits_{x to infty} Bigl(frac{2x - 1}{2x + 1}Bigr)^{x} = limlimits_{x to infty} Bigl(1 + frac{-2}{2x + 1}Bigr)^{x} = limlimits_{x to infty} Bigl(Bigl(underbrace{Bigl(1 + frac{-2}{2x + 1}Bigr)^frac{2x + 1}{-2}}_{e}Bigr)^frac{-2}{2x + 1}Bigr)^{x} = limlimits_{x to infty} e^frac{-2x}{2x + 1} =\= e^{-1}$