Question-194879

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

Answered by cortano12 last updated on 18/Jul/23

= \lim_{u \to 0} \frac{{(1 + 2 u)}^{u} - 1}{{(1 + 3 u)}^{u} - 1}

[u = \frac{1}{x}]

t h e n a p p l y L^{'} H o p i t a l

L = \lim_{x \to 0} \frac{{(1 + 2 x)}^{x} (\ln (1 + 2 x) + \frac{2}{1 + 2 x})}{{(1 + 3 x)}^{x} (\ln (1 + 3 x) + \frac{3}{1 + 3 x})}

= \begin{matrix} \frac{2}{3} \end{matrix}