اوجد نهاية

اوجد النهايه الاتيه
$\lim_{x\rightarrow 0}\frac{2x-2ln(x+1)-xln(x+1)}{xln^2(x+1)}$

$\large {\color{Red} \lim_{x\rightarrow 0}\frac{2x-xln(x+1)-2ln(x+1)}{xln^{2}(x+1)};;;let[ln(x+1)=y\rightarrow x=e^{y}-1]}$
${\color{DarkBlue} L=\lim_{y\rightarrow 0}\frac{2(e^{y}-1)-y(e^{y}-1)-2y}{y^{2}(e^{y}-1)}=\lim_{y\rightarrow 0}\frac{\frac{2(e^{y}-1)-y(e^{y}-1)-2y}{y^{3}}}{\frac{e^{y}-1}{y}}}$
${\color{DarkBlue} \lim_{y\rightarrow 0}\frac{e^{y}-1}{y}=1;;e^{y}\cong 1+y+\frac{y^{2}}{2!}+\frac{y^{3}}{3!}}$
${\color{DarkBlue} L=\lim_{y\rightarrow 0}\frac{\frac{2y^{3}}{3!}-\frac{y^{3}}{2!}}{y^{3}}=\frac{1}{3}-\frac{1}{2}=-\frac{1}{b}}$