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Question Number 1573 by 123456 last updated on 20/Aug/15
lim_(ε→+∞) ∫_0 ^ε ε^(−t) dt=?
$$\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}\:\underset{\mathrm{0}} {\overset{\epsilon} {\int}}\epsilon^{−{t}} {dt}=? \\ $$
Commented by 112358 last updated on 21/Aug/15
Let I(ε)=∫_0 ^( ε) ε^(−t) dt (ε>0) If u=−t⇒−du=dt ∴I(ε)=−∫_0 ^( −ε) ε^u du=−(ε^u /(lnε))∣_0 ^(−ε) I(ε)=−(1/(lnε))(ε^(−ε) −ε^0 ) =(1/(lnε))(1−(1/ε^ε )) I(ε)=((ε^ε −1)/(lnε^ε^ε )) ∴If L=lim_(ε→+∞) I(ε)=lim_(ε→+∞) (1/(lnε))(1−(1/ε^ε )) L=(lim_(ε→+∞) (1/(lnε)))(lim_(ε→+∞) 1−lim_(ε→+∞) (1/ε^ε )) L=(((lim_(ε→+∞) 1)/(lim_(ε→+∞) lnε)))(lim_(ε→+∞) 1−((lim_(ε→+∞) 1)/(lim_(ε→+∞) ε^ε ))) ∵ lim_(ε→+∞) 1=1 ,lim_(ε→+∞) lnε=+∞ , lim_(ε→+∞) ε^ε =+∞ ⇒((lim_(ε→+∞) 1)/(lim_(ε→+∞) lnε))=(1/(+∞))=0 lim_(ε→+∞) 1−((lim_(ε→+∞) 1)/(lim_(ε→+∞) ε^ε ))=1−(1/(+∞))=1 ∴ L=0×1=0 Proof of L=lim_(x→+∞) x^x =∞. Informally, L=lim_(x→+∞) x^x =lim_(x→+∞) e^(lnx^x ) L=lim_(x→+∞) e^(xlnx) =e^(lim_(x→+∞) xlnx) L=exp((lim_(x→0) x)(lim_(x→+∞) lnx)) lim_(x→+∞) lnx=∞, lim_(x→+∞) x=∞ ∴ L=e^(∞×∞) =∞
$${Let}\:{I}\left(\epsilon\right)=\int_{\mathrm{0}} ^{\:\epsilon} \epsilon^{−{t}} {dt}\:\:\:\:\:\:\left(\epsilon>\mathrm{0}\right) \\ $$$${If}\:{u}=−{t}\Rightarrow−{du}={dt} \\ $$$$\therefore{I}\left(\epsilon\right)=−\int_{\mathrm{0}} ^{\:−\epsilon} \epsilon^{{u}} {du}=−\frac{\epsilon^{{u}} }{{ln}\epsilon}\mid_{\mathrm{0}} ^{−\epsilon} \\ $$$${I}\left(\epsilon\right)=−\frac{\mathrm{1}}{{ln}\epsilon}\left(\epsilon^{−\epsilon} −\epsilon^{\mathrm{0}} \right) \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{ln}\epsilon}\left(\mathrm{1}−\frac{\mathrm{1}}{\epsilon^{\epsilon} }\right) \\ $$$${I}\left(\epsilon\right)=\frac{\epsilon^{\epsilon} −\mathrm{1}}{{ln}\epsilon^{\epsilon^{\epsilon} } } \\ $$$$\therefore{If}\:{L}=\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}{I}\left(\epsilon\right)=\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{ln}\epsilon}\left(\mathrm{1}−\frac{\mathrm{1}}{\epsilon^{\epsilon} }\right) \\ $$$${L}=\left(\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{ln}\epsilon}\right)\left(\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}1}−\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\epsilon^{\epsilon} }\right) \\ $$$${L}=\left(\frac{\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}1}}{\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}{ln}\epsilon}\right)\left(\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}1}−\frac{\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}1}}{\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}\epsilon^{\epsilon} }\right) \\ $$$$\because\:\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}1}=\mathrm{1}\:,\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}{ln}\epsilon=+\infty\:,\:\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}\epsilon^{\epsilon} =+\infty\:\: \\ $$$$\Rightarrow\frac{\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}1}}{\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}{ln}\epsilon}=\frac{\mathrm{1}}{+\infty}=\mathrm{0} \\ $$$$\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}1}−\frac{\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}1}}{\underset{\epsilon\rightarrow+\infty} {\mathrm{lim}}\epsilon^{\epsilon} }=\mathrm{1}−\frac{\mathrm{1}}{+\infty}=\mathrm{1} \\ $$$$\therefore\:{L}=\mathrm{0}×\mathrm{1}=\mathrm{0}\:\: \\ $$$$ \\ $$$${Proof}\:{of}\:{L}=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{x}^{{x}} =\infty.\: \\ $$$${Informally}, \\ $$$${L}=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{x}^{{x}} =\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{e}^{{lnx}^{{x}} } \\ $$$${L}=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{e}^{{xlnx}} ={e}^{\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{xlnx}} \\ $$$${L}={exp}\left(\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}\right)\left(\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{lnx}\right)\right) \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{lnx}=\infty,\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{x}=\infty \\ $$$$\therefore\:{L}={e}^{\infty×\infty} =\infty \\ $$$$ \\ $$$$ \\ $$

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