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

Calculate :: lim_(x→+∞) (((x+a)^(x+a) (x+b)^(x+b) )/((x+a+b)^(2x+a+b) ))=?

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

Answered by LEKOUMA last updated on 14/Apr/22

lim_(x→+∞) ((x^(x+a) (1+(a/x))^(x+a) x^(x+b) (1+(b/x))^(x+b) )/(x^(2x+a+b) (1+(a/x)+(b/x))^(2x+a+b) )) lim_(x→+∞) ((x^(2x+a+b) (1+(a/x))^(x+a) (1+(b/x))^(x+b) )/(x^(2x+a+b) (1+(a/x)+(b/x))^(2x+a+b) )) lim_(x→+∞) (((1+(a/x))^(x+a) (1+(b/x))^(x+b) )/((1+(a/x)+(b/x))^(2x+a+b) )) lim_(x→+∞) ((e^((x+a) (1+(a/x)−1)) e^((x+b)(1+(b/x)−1)) )/e^((2x+a+b)(1+(a/x)+(b/x)−1)) ) lim_(x→+∞) ((e^((x+a)((a/x))) e^((x+b)((b/x))) )/e^((2x+a+b)((a/x)+(b/x))) ) lim_(x→+∞) ((e^(a+(a^2 /x)) e^(b+(b^2 /x)) )/e^((2+(a/x)+(b/x))(a+b)) )=((e^a e^b )/e^(2(a+b)) )=e^(a+b) ×e^(−2(a+b)) =e^(a+b) ×e^(−2a−2b) =e^(−a−b) =e^(−(a+b))

$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{x}^{{x}+{a}} \left(\mathrm{1}+\frac{{a}}{{x}}\right)^{{x}+{a}} {x}^{{x}+{b}} \left(\mathrm{1}+\frac{{b}}{{x}}\right)^{{x}+{b}} }{{x}^{\mathrm{2}{x}+{a}+{b}} \left(\mathrm{1}+\frac{{a}}{{x}}+\frac{{b}}{{x}}\right)^{\mathrm{2}{x}+{a}+{b}} } \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{x}^{\mathrm{2}{x}+{a}+{b}} \left(\mathrm{1}+\frac{{a}}{{x}}\right)^{{x}+{a}} \left(\mathrm{1}+\frac{{b}}{{x}}\right)^{{x}+{b}} }{{x}^{\mathrm{2}{x}+{a}+{b}} \left(\mathrm{1}+\frac{{a}}{{x}}+\frac{{b}}{{x}}\right)^{\mathrm{2}{x}+{a}+{b}} } \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{{a}}{{x}}\right)^{{x}+{a}} \left(\mathrm{1}+\frac{{b}}{{x}}\right)^{{x}+{b}} }{\left(\mathrm{1}+\frac{{a}}{{x}}+\frac{{b}}{{x}}\right)^{\mathrm{2}{x}+{a}+{b}} } \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{e}^{\left({x}+{a}\right)\:\left(\mathrm{1}+\frac{{a}}{{x}}−\mathrm{1}\right)} {e}^{\left({x}+{b}\right)\left(\mathrm{1}+\frac{{b}}{{x}}−\mathrm{1}\right)} }{{e}^{\left(\mathrm{2}{x}+{a}+{b}\right)\left(\mathrm{1}+\frac{{a}}{{x}}+\frac{{b}}{{x}}−\mathrm{1}\right)} } \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{e}^{\left({x}+{a}\right)\left(\frac{{a}}{{x}}\right)} {e}^{\left({x}+{b}\right)\left(\frac{{b}}{{x}}\right)} }{{e}^{\left(\mathrm{2}{x}+{a}+{b}\right)\left(\frac{{a}}{{x}}+\frac{{b}}{{x}}\right)} } \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{e}^{{a}+\frac{{a}^{\mathrm{2}} }{{x}}} {e}^{{b}+\frac{{b}^{\mathrm{2}} }{{x}}} }{{e}^{\left(\mathrm{2}+\frac{{a}}{{x}}+\frac{{b}}{{x}}\right)\left({a}+{b}\right)} }=\frac{{e}^{{a}} {e}^{{b}} }{{e}^{\mathrm{2}\left({a}+{b}\right)} }={e}^{{a}+{b}} ×{e}^{−\mathrm{2}\left({a}+{b}\right)} \\ $$$$={e}^{{a}+{b}} ×{e}^{−\mathrm{2}{a}−\mathrm{2}{b}} ={e}^{−{a}−{b}} ={e}^{−\left({a}+{b}\right)} \\ $$

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