Chương 4: GIỚI HẠN

NL
8 tháng 3 2022 lúc 16:37

\(\lim\limits_{x\rightarrow3}\dfrac{2-\sqrt[3]{x+5}}{x^2-9}=\lim\limits_{x\rightarrow3}\dfrac{\left(2-\sqrt[3]{x+5}\right)\left(4+2\sqrt[3]{x+5}+\sqrt[3]{\left(x+5\right)^2}\right)}{\left(x-3\right)\left(x+3\right)\left(4+2\sqrt[3]{x+5}+\sqrt[3]{\left(x+5\right)^2}\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{3-x}{\left(x-3\right)\left(x+3\right)\left(4+2\sqrt[3]{x+5}+\sqrt[3]{\left(x+5\right)^2}\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{-1}{\left(x+3\right)\left(4+2\sqrt[3]{x+5}+\sqrt[3]{\left(x+5\right)^2}\right)}\)

\(=\dfrac{-1}{6\left(4+2.2+2^2\right)}=-\dfrac{1}{72}\)

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NL
8 tháng 3 2022 lúc 16:39

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{2x-1}-1}{\sqrt[]{3x+1}-2}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{2x-1}-1\right)\left(\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1\right)\left(\sqrt[]{3x+1}+2\right)}{\left(\sqrt[]{3x+1}-2\right)\left(\sqrt[]{3x+1}+2\right)\left(\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)\left(\sqrt[]{3x+1}+2\right)}{3\left(x-1\right)\left(\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{2\left(\sqrt[]{3x+1}+2\right)}{3\left(\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1\right)}\)

\(=\dfrac{2\left(2+2\right)}{3\left(1+1+1\right)}=\dfrac{8}{9}\)

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