TQ : $\mathop {\lim }\limits_{x \to 1}\frac{\sqrt{x}.\sqrt[3]{x}.\sqrt[4]{x}...\sqrt[n]{x}-1}{x-1}=\mathop {\lim }\limits_{x \to 1}\frac{(\sqrt[n]{x}-1)\sqrt[n-1]{x}...\sqrt{x}+...+(\sqrt{x}-1)}{x-1}=\mathop {\lim }\limits_{x \to 1}\frac{1}{\sqrt{x}+1}+...+\frac{1}{\sqrt[n-1]{x}+...+1}=\frac{1}{2}+...+\frac{1}{n}$Tự áp dụng :v
TQ : $\mathop {\lim }\limits_{x \to 1}\frac{\sqrt{x}.\sqrt[3]{x}.\sqrt[4]{x}...\sqrt[n]{x}-1}{x-1}=\frac{1}{2}+...+\frac{1}{n}$Tự áp dụng :v
TQ : $\mathop {\lim }\limits_{x \to 1}\frac{\sqrt{x}.\sqrt[3]{x}.\sqrt[4]{x}...\sqrt[n]{x}-1}{x-1}=\
mathop {\lim }\limits_{x \to 1}\frac{(\sqrt[n]{x}-1)\sqrt[n-1]{x}...\sqrt{x}+...+(\sqrt{x}-1)}{x-1}=\mathop {\lim }\limits_{x \to 1}\frac{1}{\sqrt{x}+1}+...+\frac{1}{\sqrt[n-1]{x}+...+1}=\frac{1}{2}+...+\frac{1}{n}$Tự áp dụng :v