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\(\begin{array}{l} \lim\limits _{x \rightarrow 0} \frac{\sqrt[n]{1+a x}-1}{x}=\lim\limits _{x \rightarrow 0} \frac{(\sqrt[n]{1+a x}-1)\left(\sqrt[n]{(1+a x)^{n-1}}+\sqrt[n]{(1+a x)^{n-2}}+\ldots+\sqrt[n]{1+a x}+1\right)}{x\left(\sqrt[n]{(1+a x)^{n-1}}+\sqrt[n]{(1+a x)^{n-2}}+\ldots+\sqrt[n]{1+a x}+1\right)} \\ \text { }=\lim\limits _{x \rightarrow 0} \frac{a}{\sqrt[n]{(1+a x)^{n-1}}+\sqrt[n]{(1+a x)^{n-2}}+\ldots+\sqrt[n]{1+a x}+1}=\frac{a}{n} \end{array}\)
\(\begin{aligned} \lim\limits _{x \rightarrow 0} \frac{x}{\sqrt[m]{1+b x}-1}=\lim\limits _{x \rightarrow 0} \frac{x\left(\sqrt[m]{(1+b x)^{m-1}}+\sqrt[m]{(1+b x)^{m-2}}+\ldots+\sqrt[m]{1+b x}+1\right)}{(\sqrt[m]{1+b x}-1)\left(\sqrt[m]{(1+b x)^{m-1}}+\sqrt[m]{(1+b x)^{m-2}}+\ldots+\sqrt[m]{1+b x}+1\right)} \\ \text { } =\lim\limits _{x \rightarrow 0} \frac{\sqrt[m]{(1+b x)^{m-1}}+\sqrt[n]{(1+b x)^{m-2}}+\ldots+\sqrt[m]{1+b x}+1}{b}=\frac{m}{b} \end{aligned}\)
Khi đó
\(A=\lim\limits _{x \rightarrow 0} \frac{(\sqrt[n]{1+a x}-1) x}{x(\sqrt[m]{1+b x}-1)}=\lim\limits _{x \rightarrow 0} \frac{\sqrt[n]{1+a x}-1}{x} \cdot \lim\limits _{x \rightarrow 0} \frac{x}{\sqrt[n]{1+b x}-1}=\frac{a}{n} \cdot \frac{m}{b}=\frac{a m}{b n}\)