$\lim\limits_{x\to0}\dfrac{x^2}{\sqrt{1+2x}-\sqrt[3]{1+3x}}$ tính nha 13/10/2021 Bởi Margaret $\lim\limits_{x\to0}\dfrac{x^2}{\sqrt{1+2x}-\sqrt[3]{1+3x}}$ tính nha
$\lim\limits_{x\to 0}\dfrac{x^2}{ \sqrt{1+2x}-1+1-\sqrt[3]{1+3x} }$ $=\lim\limits_{x\to 0}\dfrac{x^2}{ \dfrac{1+2x-1}{\sqrt{1+2x}+1} + \dfrac{1-1-3x}{ \sqrt[3]{1+3x}^2+\sqrt[3]{1+3x}+1} }$ $=\lim\limits_{x\to 0}\dfrac{x}{ \dfrac{2}{\sqrt{1+2x}+1} -\dfrac{3}{\sqrt[3]{1+3x}^2+\sqrt[3]{1+3x}+ 1} }$ $=\lim\limits_{x\to 0}\dfrac{x}{ \dfrac{2}{\sqrt{2x+1}+1}-\dfrac{ 3}{(3x+1)^{\frac{2}{3}} +(3x+1)^{\frac{1}{3}}+1} }$ $=\lim\limits_{x\to 0}\dfrac{1}{ \dfrac{-2(\sqrt{2x+1}+1)’}{ (\sqrt{2x+1}+1)^2}+ \dfrac{3[ (3x+1)^{\frac{2}{3}}+(3x+1)^{\frac{1}{3}}+1]’ }{ [(3x+1)^{\frac{2}{3}}+(3x+1)^{\frac{1}{3}}+1]^2 }}$ $=\lim\limits_{x\to 0}\dfrac{1}{ \dfrac{-2}{\sqrt{2x+1}(\sqrt{2x+1}+1)^2} +\dfrac{3[ 2(3x+1)^{\frac{-1}{3}}+ (3x+1)^{\frac{-2}{3}} ] }{ [(3x+1)^{\frac{2}{3}}+(3x+1)^{\frac{1}{3}}+1]^2}}$ $=2$ Bình luận
`text{Cách 2 đang được cập nhật}`
$\lim\limits_{x\to 0}\dfrac{x^2}{ \sqrt{1+2x}-1+1-\sqrt[3]{1+3x} }$
$=\lim\limits_{x\to 0}\dfrac{x^2}{ \dfrac{1+2x-1}{\sqrt{1+2x}+1} + \dfrac{1-1-3x}{ \sqrt[3]{1+3x}^2+\sqrt[3]{1+3x}+1} }$
$=\lim\limits_{x\to 0}\dfrac{x}{ \dfrac{2}{\sqrt{1+2x}+1} -\dfrac{3}{\sqrt[3]{1+3x}^2+\sqrt[3]{1+3x}+ 1} }$
$=\lim\limits_{x\to 0}\dfrac{x}{ \dfrac{2}{\sqrt{2x+1}+1}-\dfrac{ 3}{(3x+1)^{\frac{2}{3}} +(3x+1)^{\frac{1}{3}}+1} }$
$=\lim\limits_{x\to 0}\dfrac{1}{ \dfrac{-2(\sqrt{2x+1}+1)’}{ (\sqrt{2x+1}+1)^2}+ \dfrac{3[ (3x+1)^{\frac{2}{3}}+(3x+1)^{\frac{1}{3}}+1]’ }{ [(3x+1)^{\frac{2}{3}}+(3x+1)^{\frac{1}{3}}+1]^2 }}$
$=\lim\limits_{x\to 0}\dfrac{1}{ \dfrac{-2}{\sqrt{2x+1}(\sqrt{2x+1}+1)^2} +\dfrac{3[ 2(3x+1)^{\frac{-1}{3}}+ (3x+1)^{\frac{-2}{3}} ] }{ [(3x+1)^{\frac{2}{3}}+(3x+1)^{\frac{1}{3}}+1]^2}}$
$=2$