tinhs $\lim\limits{x\to 0} \dfrac{\sqrt[3]{x+1}-\sqrt[3]{x^3+1}}{x}$ 25/10/2021 Bởi Caroline tinhs $\lim\limits{x\to 0} \dfrac{\sqrt[3]{x+1}-\sqrt[3]{x^3+1}}{x}$
Đáp án: `1/3` Giải thích các bước giải: $\lim\limits_{x\to0}\dfrac{\sqrt[3]{x+1}-\sqrt[3]{x^3+1}}{x}=\lim\limits_{x\to0}\dfrac{(x+1-x^3-1)}{x.(\sqrt[3]{x+1}^2+\sqrt[3]{(x+1)(x^3+1)}+\sqrt[3]{x^3+1}^2)}= \lim\limits_{x\to0}\dfrac{1-x^2} {\sqrt[3]{x+1}^2+\sqrt[3]{(x+1)(x^3+1)}+\sqrt[3]{x^3+1}^2}= \dfrac{1}{3}$ Bình luận
`~rai~` \(\lim\limits_{x\to 0}\dfrac{\sqrt[3]{x+1}-\sqrt[3]{x^3+1}}{x}\\=\lim\limits_{x\to 0}\dfrac{\left(\sqrt[3]{x+1}-\sqrt[3]{x^3+1}\right)\left(\sqrt[3]{x+1}^2+\sqrt[3]{\left(x+1\right)\left(x^3+1\right)}+\sqrt{x^3+1}^2\right)}{x\left(\sqrt[3]{x+1}^2+\sqrt[3]{\left(x+1\right)\left(x^3+1\right)}+\sqrt[3]{x^3+1}^2\right)}\\=\lim\limits_{x\to 0}\dfrac{x+1-x^3-1}{x\left(\sqrt[3]{x+1}^2+\sqrt[3]{\left(x+1\right)\left(x^3+1\right)}+\sqrt[3]{x^3+1}^2\right)}\\=\lim\limits_{x\to 0}\dfrac{1-x^2}{\sqrt[3]{x+1}^2+\sqrt[3]{\left(x+1\right)\left(x^3+1\right)}+\sqrt[3]{x^3+1}^2}\\\\=\dfrac{1-0^2}{\sqrt[3]{0+1}^2+\sqrt[3]{\left(0+1\right)\left(0^3+1\right)}+\sqrt[3]{0^3+1}}\\=\dfrac{1}{3}\) Bình luận
Đáp án:
`1/3`
Giải thích các bước giải:
$\lim\limits_{x\to0}\dfrac{\sqrt[3]{x+1}-\sqrt[3]{x^3+1}}{x}=\lim\limits_{x\to0}\dfrac{(x+1-x^3-1)}{x.(\sqrt[3]{x+1}^2+\sqrt[3]{(x+1)(x^3+1)}+\sqrt[3]{x^3+1}^2)}= \lim\limits_{x\to0}\dfrac{1-x^2} {\sqrt[3]{x+1}^2+\sqrt[3]{(x+1)(x^3+1)}+\sqrt[3]{x^3+1}^2}= \dfrac{1}{3}$
`~rai~`
\(\lim\limits_{x\to 0}\dfrac{\sqrt[3]{x+1}-\sqrt[3]{x^3+1}}{x}\\=\lim\limits_{x\to 0}\dfrac{\left(\sqrt[3]{x+1}-\sqrt[3]{x^3+1}\right)\left(\sqrt[3]{x+1}^2+\sqrt[3]{\left(x+1\right)\left(x^3+1\right)}+\sqrt{x^3+1}^2\right)}{x\left(\sqrt[3]{x+1}^2+\sqrt[3]{\left(x+1\right)\left(x^3+1\right)}+\sqrt[3]{x^3+1}^2\right)}\\=\lim\limits_{x\to 0}\dfrac{x+1-x^3-1}{x\left(\sqrt[3]{x+1}^2+\sqrt[3]{\left(x+1\right)\left(x^3+1\right)}+\sqrt[3]{x^3+1}^2\right)}\\=\lim\limits_{x\to 0}\dfrac{1-x^2}{\sqrt[3]{x+1}^2+\sqrt[3]{\left(x+1\right)\left(x^3+1\right)}+\sqrt[3]{x^3+1}^2}\\\\=\dfrac{1-0^2}{\sqrt[3]{0+1}^2+\sqrt[3]{\left(0+1\right)\left(0^3+1\right)}+\sqrt[3]{0^3+1}}\\=\dfrac{1}{3}\)