Question Number 205451 by MrGHK last updated on 21/Mar/24
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{{e}^{\mathrm{2}{t}} {t}} \\ $$
Answered by Berbere last updated on 21/Mar/24
![x→^f e^(−2x) ;over [0,t] t≤(1/2);∃c∈]0,t[ Such ⇒f(t)−f(0)=tf′(c) ⇒e^(−2t) −1=−2te^(−2c) ;e^(−2c) ≤0⇒e^(−2t) −1≥−2t ⇒e^(−2t) ≥1−2t ∫_0 ^x (dt/(e^(2t) .t))=f(x);f′(x)>0 f increse ⇒(lim_(x→∞) f(x)≥f(a);∀a∈R_+ ) lim_(x→∞) f(x)≥f((1/2))=∫_0 ^(1/2) (e^(−2t) /t)dt≥∫_0 ^(1/2) (1/t)(1−2t)dt =[ln(t)−2]_0 ^(1/2) =+∞ lim_(x→∞) f(x)=+∞ or just saying;(1/(e^(−2x) x))∼^0 (1/x) diverge](Q205469.png)
$$\left.{x}\overset{{f}} {\rightarrow}{e}^{−\mathrm{2}{x}} ;{over}\:\left[\mathrm{0},{t}\right]\:{t}\leqslant\frac{\mathrm{1}}{\mathrm{2}};\exists{c}\in\right]\mathrm{0},{t}\left[\:{Such}\right. \\ $$$$\Rightarrow{f}\left({t}\right)−{f}\left(\mathrm{0}\right)={tf}'\left({c}\right) \\ $$$$\Rightarrow{e}^{−\mathrm{2}{t}} −\mathrm{1}=−\mathrm{2}{te}^{−\mathrm{2}{c}} ;{e}^{−\mathrm{2}{c}} \leqslant\mathrm{0}\Rightarrow{e}^{−\mathrm{2}{t}} −\mathrm{1}\geqslant−\mathrm{2}{t} \\ $$$$\Rightarrow{e}^{−\mathrm{2}{t}} \geqslant\mathrm{1}−\mathrm{2}{t}\: \\ $$$$\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{{e}^{\mathrm{2}{t}} .{t}}={f}\left({x}\right);{f}'\left({x}\right)>\mathrm{0}\:{f}\:{increse}\:\Rightarrow\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}\right)\geqslant{f}\left({a}\right);\forall{a}\in\mathbb{R}_{+} \right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}\right)\geqslant{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{e}^{−\mathrm{2}{t}} }{{t}}{dt}\geqslant\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{{t}}\left(\mathrm{1}−\mathrm{2}{t}\right){dt} \\ $$$$=\left[{ln}\left({t}\right)−\mathrm{2}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} =+\infty \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}\right)=+\infty \\ $$$${or}\:{just}\:{saying};\frac{\mathrm{1}}{{e}^{−\mathrm{2}{x}} {x}}\overset{\mathrm{0}} {\sim}\frac{\mathrm{1}}{{x}}\:{diverge} \\ $$