Tìm giới hạn \(N=\lim\limits _{x \rightarrow 0} \frac{\sqrt[m]{1+a x}-\sqrt[n]{1+b x}}{\sqrt{1+x}-1}\)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} \lim \limits_{x \rightarrow 0} \frac{\sqrt[m]{1+a x}-1}{x}=\lim\limits _{x \rightarrow 0} \frac{(\sqrt[m]{1+a x}-1)\left(\sqrt[m]{(1+a x)^{m-1}}+\sqrt[m]{(1+a x)^{m-2}}+\ldots+\sqrt[m]{1+a x}+1\right)}{x\left(\sqrt[m]{(1+a x)^{m-1}}+\sqrt[m]{(1+a x)^{m-2}}+\ldots+\sqrt[m]{1+a x}+1\right)} \\ \quad=\lim \limits_{x \rightarrow 0} \frac{a}{\sqrt[m]{(1+a x)^{m-1}}+\sqrt[m]{(1+a x)^{m-2}+\ldots+\sqrt[m]{1+a x}+1}}=\frac{a}{m} \end{array}\)
\(\begin{array}{l} \lim\limits _{x \rightarrow 0} \frac{\sqrt[n]{1+b x}-1}{x}=\lim\limits _{x \rightarrow 0} \frac{(\sqrt[n]{1+b x}-1)\left(\sqrt[n]{(1+b x)^{n-1}}+\sqrt[n]{(1+b x)^{n-2}}+\ldots+\sqrt[n]{1+b x}+1\right)}{x\left(\sqrt[n]{(1+b x)^{n-1}}+\sqrt[n]{(1+b x)^{n-2}}+\ldots+\sqrt[n]{1+b x}+1\right)} \\ \quad=\lim\limits _{x \rightarrow 0} \frac{b}{\sqrt[n]{(1+b x)^{n-1}}+\sqrt[n]{(1+b x)^{n-2}+\ldots+\sqrt[n]{1+b x}+1}}=\frac{b}{n} \end{array}\)
\(N=\lim\limits _{x \rightarrow 0}\left(\frac{\sqrt[m]{1+a x}-1}{x}-\frac{\sqrt[n]{1+b x}-1}{x}\right) \cdot \frac{x}{\sqrt{1+x}-1}=\left(\frac{a}{m}-\frac{b}{n}\right) \cdot 2=\frac{2(a n-b m)}{m n}\)