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Question Number 194879 by tri26112004 last updated on 18/Jul/23

Answered by cortano12 last updated on 18/Jul/23

$$=\underset{u\to 0}{\lim }\frac{{(1+2u)}^u-1}{{(1+3u)}^u-1}$$$$[u=\frac{1}{x}]$$$$thenapply{L}^{\prime }Hopital$$$$L=\underset{x\to 0}{\lim }\frac{{(1+2x)}^x(\ln (1+2x)+\frac{2}{1+2x})}{{(1+3x)}^x(\ln (1+3x)+\frac{3}{1+3x})}$$$$=\overline{)\begin{array}{c}\frac{2}{3}\end{array}}$$$$$$

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