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Question Number 94659 by msup by abdo last updated on 20/May/20
$${calculate}\:{lim}_{{n}\rightarrow+\infty} \int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}} \:{e}^{−\mathrm{3}{t}} \:{dt} \\ $$
Answered by mathmax by abdo last updated on 20/May/20
![let A_n =∫_0 ^∞ (1−(t/n))^n e^(−3t) dt ⇒A_n =∫_R (1−(t/n))^n e^(−3t) χ_([0,+∞[) (t)dt =∫_R f_n (t) dt we have f_n →^(cs) f =e^(−4t) and ∣f_n ∣≤e^(−4t) ⇒ theorem of convergence dominee give lim_(n→+∞) A_n =∫_R limf_n (t)dt =∫_0 ^∞ e^(−4t) dt =[−(1/4)e^(−4t) ]_0 ^(+∞) =(1/4)](Q94709.png)
$$\mathrm{let}\:\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{n}}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{3t}} \:\mathrm{dt}\:\Rightarrow\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{R}} \left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{n}}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{3t}} \:\chi_{\left[\mathrm{0},+\infty\left[\right.\right.} \:\left(\mathrm{t}\right)\mathrm{dt} \\ $$$$=\int_{\mathrm{R}} \mathrm{f}_{\mathrm{n}} \left(\mathrm{t}\right)\:\mathrm{dt}\:\:\mathrm{we}\:\mathrm{have}\:\:\mathrm{f}_{\mathrm{n}} \rightarrow^{\mathrm{cs}} \:\:\mathrm{f}\:=\mathrm{e}^{−\mathrm{4t}} \:\:\:\:\mathrm{and}\:\mid\mathrm{f}_{\mathrm{n}} \mid\leqslant\mathrm{e}^{−\mathrm{4t}} \:\Rightarrow \\ $$$$\mathrm{theorem}\:\mathrm{of}\:\mathrm{convergence}\:\mathrm{dominee}\:\mathrm{give} \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{R}} \mathrm{limf}_{\mathrm{n}} \left(\mathrm{t}\right)\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{4t}} \:\mathrm{dt}\:=\left[−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{e}^{−\mathrm{4t}} \right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$