Tìm giới hạn \(A=\lim\limits _{x \rightarrow 0} \frac{\sqrt[n]{1+a x}-1}{\sqrt[m]{1+b x}-1} \text { với } a b \neq 0:\)
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Lời giải:
Báo sai\(\begin{array}{l} A = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + ax}} - 1}}{{\sqrt[m]{{1 + bx}} - 1}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + ax}} - 1}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt[m]{{1 + bx}} - 1}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{(\sqrt[n]{{1 + ax}} - 1)\left( {\sqrt[n]{{{{(1 + ax)}^{n - 1}}}} + \sqrt[n]{{{{(1 + ax)}^{n - 2}}}} + \ldots + \sqrt[n]{{1 + ax}} + 1} \right)}}{{x\left( {\sqrt[n]{{{{(1 + ax)}^{n - 1}}}} + \sqrt[n]{{{{(1 + ax)}^{n - 2}}}} + \ldots + \sqrt[n]{{1 + ax}} + 1} \right)}}.\mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt[m]{{{{(1 + bx)}^{m - 1}}}} + \sqrt[m]{{{{(1 + bx)}^{m - 2}}}} + \ldots + \sqrt[m]{{1 + bx}} + 1} \right)}}{{(\sqrt[m]{{1 + bx}} - 1)\left( {\sqrt[m]{{{{(1 + bx)}^{m - 1}}}} + \sqrt[m]{{{{(1 + bx)}^{m - 2}}}} + \ldots + \sqrt[m]{{1 + bx}} + 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{a}{{\sqrt[n]{{{{(1 + ax)}^{n - 1}}}} + \sqrt[n]{{{{(1 + ax)}^{n - 2}}}} + \ldots + \sqrt[n]{{1 + ax}} + 1}}.\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{{{(1 + bx)}^{m - 1}}}} + \sqrt[m]{{{{(1 + bx)}^{m - 2}}}} + \ldots + \sqrt[m]{{1 + bx}} + 1}}{b}\\ = \frac{a}{n} \cdot \frac{m}{b} = \frac{{am}}{{bn}} \end{array}\)
