tìm giới hạn của hàm số

D

demon311

$\lim \limits_{x \to 0} \dfrac{ (1-2x)\sqrt[3]{ 1+3x}-1}{x} = \lim \limits_{x \to 0} \left ( \sqrt[3]{ \dfrac{ 1}{x^3}+\dfrac{ 3}{x^2}}-2\sqrt[]{ 1+3x}-\dfrac{ 1}{x} \right ) \\
\lim \limits_{x \to 0} \left ( \sqrt[3]{ \dfrac{ 1}{x^3}+\dfrac{ 3}{x^2}}-\dfrac{ 1}{x} \right ) = \lim \limits_{x \to 0} \dfrac{ \dfrac{ 3}{x^2}}{\sqrt[3]{ \left (\dfrac{ 1}{x^3}+\dfrac{ 3}{x^2} \right )^2} +\dfrac{ 1}{x^2}+\sqrt[3]{ \dfrac{ 1}{x^3}+\dfrac{ 3}{x^2}}.\dfrac{ 1}{x}} =
\lim \limits_{x \to 0} \dfrac{ 3}{1+1+1}=1 \\
\Rightarrow \lim \limits_{x \to 0} \dfrac{ (1-2x)\sqrt[3]{ 1+3x}-1}{x} = 1-2=-1
$

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