Question Number 194879 by tri26112004 last updated on 18/Jul/23
Answered by cortano12 last updated on 18/Jul/23
![= lim_(u→0) (((1+2u)^u −1)/((1+3u)^u −1)) [ u =(1/x) ] then apply L′Hopital L= lim_(x→0) (((1+2x)^x (ln (1+2x)+(2/(1+2x))))/((1+3x)^x (ln (1+3x)+(3/(1+3x))))) = determinant (((2/3)))](https://www.tinkutara.com/question/Q194885.png)
$$\:\:=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{2}{u}\right)^{{u}} −\mathrm{1}}{\left(\mathrm{1}+\mathrm{3}{u}\right)^{{u}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\left[\:{u}\:=\frac{\mathrm{1}}{{x}}\:\right] \\ $$$$\:{then}\:{apply}\:{L}'{Hopital}\: \\ $$$$\:\:{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{2}{x}\right)^{{x}} \left(\mathrm{ln}\:\left(\mathrm{1}+\mathrm{2}{x}\right)+\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}{x}}\right)}{\left(\mathrm{1}+\mathrm{3}{x}\right)^{{x}} \left(\mathrm{ln}\:\left(\mathrm{1}+\mathrm{3}{x}\right)+\frac{\mathrm{3}}{\mathrm{1}+\mathrm{3}{x}}\right)} \\ $$$$\:\:=\:\begin{array}{|c|}{\frac{\mathrm{2}}{\mathrm{3}}}\\\hline\end{array} \\ $$$$ \\ $$