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Question Number 28098 by tawa tawa last updated on 20/Jan/18

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{\log }_{\mathrm{e}}\mathrm{x}}{{\mathrm{x}}^{\mathrm{h}}},\mathrm{h}>0$$

Answered by ajfour last updated on 20/Jan/18

$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{\left(1/x\right)}{{hx}^{h-1}}=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{{hx}^h}=0.$$

Commented bytawa tawa last updated on 21/Jan/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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