Toán lim (x->1) $\frac{\sqrt[m]{x} – 1}{\sqrt[n]{x} – 1 }$ 03/12/2021 By Valentina lim (x->1) $\frac{\sqrt[m]{x} – 1}{\sqrt[n]{x} – 1 }$
Ta có $\underset{x \to 1}{\lim} \dfrac{\sqrt[m]{x} – 1}{\sqrt[n]{x} – 1} = \underset{x \to 1}{\lim} \dfrac{(x-1)(\sqrt[n]{x^{n-1}} + \sqrt[n]{x^{n-2}} + \cdots + \sqrt[n]{x} + 1)}{(x-1)(\sqrt[m]{x^{n-1}} + \sqrt[m]{x^{n-2}} + \cdots + \sqrt[m]{x} + 1)}$ $= \underset{x \to 1}{\lim} \dfrac{\sqrt[n]{x^{n-1}} + \sqrt[n]{x^{n-2}} + \cdots + \sqrt[n]{x} + 1}{\sqrt[m]{x^{n-1}} + \sqrt[m]{x^{n-2}} + \cdots + \sqrt[m]{x} + 1}$ $= \dfrac{n}{m}$ Vậy $\underset{x \to 1}{\lim} \dfrac{\sqrt[m]{x} – 1}{\sqrt[n]{x} – 1} = \dfrac{n}{m}$. Reply
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Ta có
$\underset{x \to 1}{\lim} \dfrac{\sqrt[m]{x} – 1}{\sqrt[n]{x} – 1} = \underset{x \to 1}{\lim} \dfrac{(x-1)(\sqrt[n]{x^{n-1}} + \sqrt[n]{x^{n-2}} + \cdots + \sqrt[n]{x} + 1)}{(x-1)(\sqrt[m]{x^{n-1}} + \sqrt[m]{x^{n-2}} + \cdots + \sqrt[m]{x} + 1)}$
$= \underset{x \to 1}{\lim} \dfrac{\sqrt[n]{x^{n-1}} + \sqrt[n]{x^{n-2}} + \cdots + \sqrt[n]{x} + 1}{\sqrt[m]{x^{n-1}} + \sqrt[m]{x^{n-2}} + \cdots + \sqrt[m]{x} + 1}$
$= \dfrac{n}{m}$
Vậy
$\underset{x \to 1}{\lim} \dfrac{\sqrt[m]{x} – 1}{\sqrt[n]{x} – 1} = \dfrac{n}{m}$.