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Question Number 168575 by qaz last updated on 13/Apr/22

$$\mathrm{Calculate}::\underset{\mathrm{x}\to +\mathrm{\infty }}{\lim }\frac{{(\mathrm{x}+\mathrm{a})}^{\mathrm{x}+\mathrm{a}}{(\mathrm{x}+\mathrm{b})}^{\mathrm{x}+\mathrm{b}}}{{(\mathrm{x}+\mathrm{a}+\mathrm{b})}^{2\mathrm{x}+\mathrm{a}+\mathrm{b}}}=?$$

Answered by LEKOUMA last updated on 14/Apr/22

$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{x^{x+a}{(1+\frac{a}{x})}^{x+a}{x}^{x+b}{(1+\frac{b}{x})}^{x+b}}{x^{2x+a+b}{(1+\frac{a}{x}+\frac{b}{x})}^{2x+a+b}}$$$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{x^{2x+a+b}{(1+\frac{a}{x})}^{x+a}{(1+\frac{b}{x})}^{x+b}}{x^{2x+a+b}{(1+\frac{a}{x}+\frac{b}{x})}^{2x+a+b}}$$$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{{(1+\frac{a}{x})}^{x+a}{(1+\frac{b}{x})}^{x+b}}{{(1+\frac{a}{x}+\frac{b}{x})}^{2x+a+b}}$$$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{e^{(x+a)(1+\frac{a}{x}-1)}{e}^{(x+b)(1+\frac{b}{x}-1)}}{e^{(2x+a+b)(1+\frac{a}{x}+\frac{b}{x}-1)}}$$$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{e^{(x+a)\left(\frac{a}{x}\right)}{e}^{(x+b)\left(\frac{b}{x}\right)}}{e^{(2x+a+b)(\frac{a}{x}+\frac{b}{x})}}$$$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{e^{a+\frac{a^2}{x}}{e}^{b+\frac{b^2}{x}}}{e^{(2+\frac{a}{x}+\frac{b}{x})(a+b)}}=\frac{e^a{e}^b}{e^{2(a+b)}}=e^{a+b}\times e^{-2(a+b)}$$$$=e^{a+b}\times e^{-2a-2b}=e^{-a-b}=e^{-(a+b)}$$

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