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Question Number 207339 by Ghisom last updated on 12/May/24

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{(x+a)}^{1/x}-x^{1/x}}{{(x+b)}^{1/x}-x^{1/x}}=?$$

Answered by sniper237 last updated on 12/May/24

$$\frac{a}{b}cause\stackrel{X=1/x}{=}\underset{X\to 0}{\lim }\frac{{(1+aX)}^X-1}{{(1+bX)}^X-1}$$$$asXln(1+aX)\sim {aX}^2$$$$then{(1+aX)}^X\approx e^{{aX}^2}\sim 1+{aX}^2$$

Commented by Ghisom last updated on 13/May/24

$$\mathrm{thank}\mathrm{you}$$

Answered by mathzup last updated on 12/May/24

$$l\left(x\right)=\frac{x^{\frac{1}{x}}({(1+\frac{a}{x})}^{\frac{1}{x}}-1)}{x^{\frac{1}{x}}({(1+\frac{b}{x})}^{\frac{1}{x}}-1)}$$$$wehave{(1+\frac{a}{x})}^{\frac{1}{x}}-1=e^{\frac{1}{x}ln(1+\frac{a}{x})}$$$$\sim e^{\frac{1}{x}\left(\frac{a}{x}\right)}-1=e^{\frac{a}{x^2}}-1\sim \frac{a}{x^2}$$$${(1+\frac{b}{x})}^{\frac{1}{x}}-1\sim \frac{b}{x^2}\Rightarrow$$$${lim}_{x\to +\mathrm{\infty }}l\left(x\right)=\frac{a}{b}(b\ne 0)$$$$$$

Commented by Ghisom last updated on 13/May/24

$$\mathrm{thank}\mathrm{you}$$

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