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lim-x-0-x-2x-ln-2-t-t-dt-ln2-2-




Question Number 95323 by ~blr237~ last updated on 24/May/20
lim_(x→0)  ∫_x ^(2x)  ((ln(2+t))/t)dt = (ln2)^2
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\int_{{x}} ^{\mathrm{2}{x}} \:\frac{{ln}\left(\mathrm{2}+{t}\right)}{{t}}{dt}\:=\:\left({ln}\mathrm{2}\right)^{\mathrm{2}} \\ $$
Answered by abdomathmax last updated on 24/May/20
∃ c∈]x,2x[ / ∫_x ^(2x)  ((ln(2+t))/t)dt =ln(2+c)∫_x ^(2x)  (dt/t)  =ln(2+c)ln(((2x)/x))   (x→0 ⇒c→0 ⇒  lim_(x→0)  ∫_x ^(2x)  ((ln(2+t))/t)dt =ln(2).ln(2) =(ln2)^2
$$\left.\exists\:\mathrm{c}\in\right]\mathrm{x},\mathrm{2x}\left[\:/\:\int_{\mathrm{x}} ^{\mathrm{2x}} \:\frac{\mathrm{ln}\left(\mathrm{2}+\mathrm{t}\right)}{\mathrm{t}}\mathrm{dt}\:=\mathrm{ln}\left(\mathrm{2}+\mathrm{c}\right)\int_{\mathrm{x}} ^{\mathrm{2x}} \:\frac{\mathrm{dt}}{\mathrm{t}}\right. \\ $$$$=\mathrm{ln}\left(\mathrm{2}+\mathrm{c}\right)\mathrm{ln}\left(\frac{\mathrm{2x}}{\mathrm{x}}\right)\:\:\:\left(\mathrm{x}\rightarrow\mathrm{0}\:\Rightarrow\mathrm{c}\rightarrow\mathrm{0}\:\Rightarrow\right. \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\int_{\mathrm{x}} ^{\mathrm{2x}} \:\frac{\mathrm{ln}\left(\mathrm{2}+\mathrm{t}\right)}{\mathrm{t}}\mathrm{dt}\:=\mathrm{ln}\left(\mathrm{2}\right).\mathrm{ln}\left(\mathrm{2}\right)\:=\left(\mathrm{ln2}\right)^{\mathrm{2}} \\ $$

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