tìm giới hạn: $\lim_{x \to 0} $\frac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}$ 16/10/2021 Bởi Everleigh tìm giới hạn: $\lim_{x \to 0} $\frac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}$
Ta có $\underset{x \to 0}{\lim} \dfrac{2\sqrt{x+1} – \sqrt[3]{8 – x}}{x} = \underset{x \to 0}{\lim} \dfrac{2(\sqrt{x+1} – 1) – (\sqrt[3]{8-x} – 2)}{x}$ $= \underset{x \to 0}{\lim} \dfrac{ \frac{2x}{\sqrt{x+1} + 1} – \frac{8-x-8}{\sqrt[3]{(8-x)^2} + 2\sqrt[3]{8-x} + 4}}{x}$ $= \underset{x \to 0}{\lim} \dfrac{2}{\sqrt{x+1} + 1} + \dfrac{1}{\sqrt[3]{(8-x)^2} + 2\sqrt[3]{8-x} + 4}$ $= \dfrac{2}{1 + 1} + \dfrac{1}{4 + 4 + 4} = \dfrac{13}{12}$ Vậy giới hạn trên bằng $\dfrac{13}{12}$. Bình luận
Đáp án: \[\lim\limits_{x\to 0}\dfrac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}=\dfrac{13}{12}\] Giải thích các bước giải: \(\lim\limits_{x\to 0}\dfrac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}\\=\lim_{x\to 0}\dfrac{\left(2\sqrt{x+1}-2\right)+\left(2-\sqrt[3]{8-x}\right)}{x}\) \(=\lim\limits_{x\to 0}\left(\dfrac{2\left(\sqrt{x+1}-1\right)}{x}+\dfrac{2-\sqrt[3]{8-x}}{x}\right)\) \(=\lim\limits_{x\to 0}\left(\dfrac{2x}{x\left(\sqrt{x+1}+1\right)}+\dfrac{8-(8-x)}{x\left(4+2\sqrt[3]{8-x}+\sqrt[3]{(8-x)^2}\right)}\right)\) \(=\lim\limits_{x\to 0}\left(\dfrac 2{\sqrt{x+1}+1}+\dfrac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\right)\) \(=\dfrac 2{\sqrt{0+1}+1}+\dfrac{1}{4+2\sqrt[3]{8-0}+\sqrt[3]{(8-0)^2}}\\ =1+\dfrac{1}{13}\) \(=\dfrac{13}{12}\) Vậy \(\lim\limits_{x\to 0}\dfrac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}=\dfrac{13}{12}\) Bình luận
Ta có
$\underset{x \to 0}{\lim} \dfrac{2\sqrt{x+1} – \sqrt[3]{8 – x}}{x} = \underset{x \to 0}{\lim} \dfrac{2(\sqrt{x+1} – 1) – (\sqrt[3]{8-x} – 2)}{x}$
$= \underset{x \to 0}{\lim} \dfrac{ \frac{2x}{\sqrt{x+1} + 1} – \frac{8-x-8}{\sqrt[3]{(8-x)^2} + 2\sqrt[3]{8-x} + 4}}{x}$
$= \underset{x \to 0}{\lim} \dfrac{2}{\sqrt{x+1} + 1} + \dfrac{1}{\sqrt[3]{(8-x)^2} + 2\sqrt[3]{8-x} + 4}$
$= \dfrac{2}{1 + 1} + \dfrac{1}{4 + 4 + 4} = \dfrac{13}{12}$
Vậy giới hạn trên bằng $\dfrac{13}{12}$.
Đáp án:
\[\lim\limits_{x\to 0}\dfrac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}=\dfrac{13}{12}\]
Giải thích các bước giải:
\(\lim\limits_{x\to 0}\dfrac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}\\=\lim_{x\to 0}\dfrac{\left(2\sqrt{x+1}-2\right)+\left(2-\sqrt[3]{8-x}\right)}{x}\)
\(=\lim\limits_{x\to 0}\left(\dfrac{2\left(\sqrt{x+1}-1\right)}{x}+\dfrac{2-\sqrt[3]{8-x}}{x}\right)\)
\(=\lim\limits_{x\to 0}\left(\dfrac{2x}{x\left(\sqrt{x+1}+1\right)}+\dfrac{8-(8-x)}{x\left(4+2\sqrt[3]{8-x}+\sqrt[3]{(8-x)^2}\right)}\right)\)
\(=\lim\limits_{x\to 0}\left(\dfrac 2{\sqrt{x+1}+1}+\dfrac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\right)\)
\(=\dfrac 2{\sqrt{0+1}+1}+\dfrac{1}{4+2\sqrt[3]{8-0}+\sqrt[3]{(8-0)^2}}\\ =1+\dfrac{1}{13}\)
\(=\dfrac{13}{12}\)
Vậy \(\lim\limits_{x\to 0}\dfrac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}=\dfrac{13}{12}\)