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Question Number 121410 by john santu last updated on 07/Nov/20

$$\underset{x\to 1}{\lim }\frac{\sqrt[\mathrm{p}]{\mathrm{x}}-1}{\sqrt[\mathrm{q}]{\mathrm{x}}-1}?$$

Answered by TANMAY PANACEA last updated on 07/Nov/20

$$x=t^{pq}$$$$li\underset{t\to 1}{m}\frac{t^q-1}{t^p-1}$$$$li\underset{t\to 1}{m}\frac{\frac{t^q-1}{t-1}}{\frac{t^p-1}{t-1}}=\frac{q\times {\left(1\right)}^{q-1}}{p\times {\left(1\right)}^{p-1}}=\frac{q}{p}$$

Answered by liberty last updated on 07/Nov/20

$$\underset{x\to 1}{\lim }\frac{{\mathrm{x}}^{\frac{1}{\mathrm{p}}}-1}{{\mathrm{x}}^{\frac{1}{\mathrm{q}}}-1}=\underset{x\to 1}{\lim }\frac{\frac{1}{\mathrm{p}}{\mathrm{x}}^{\frac{1}{\mathrm{p}}-1}}{\frac{1}{\mathrm{q}}{\mathrm{x}}^{\frac{1}{\mathrm{q}}-1}}$$$$\underset{x\to 1}{\lim }\frac{{\mathrm{qx}}^{\frac{1-\mathrm{p}}{\mathrm{p}}}}{{\mathrm{px}}^{\frac{1-\mathrm{q}}{\mathrm{q}}}}=\frac{\mathrm{q}}{\mathrm{p}}.$$

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