a.
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4+x}-2}{4x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{4+x}-2\right)\left(\sqrt{4+x}+2\right)}{4x\left(\sqrt{4+x}+2\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{x}{4x\left(\sqrt{4+x}+2\right)}=\lim\limits_{x\rightarrow0}\dfrac{1}{4\left(\sqrt{4+x}+2\right)}=\dfrac{1}{4\left(\sqrt{4+0}+2\right)}=\dfrac{1}{16}\)
b.
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{x+7}-2\right)\left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}{\left(x-1\right)\left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}\)
\(=\dfrac{1}{\sqrt[3]{8^2}+2\sqrt[3]{8}+4}=\dfrac{1}{12}\)
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