Question 35 page 270 - Mood Graybill and Boes (1974).
Finding the asymptotic distribution of $F\left( x \right)=\left( 1-{{e}^{-{x}/{\left( 1-x \right)}\;}} \right){{I}_{\left( 0,1 \right)}}\left( x \right)+{{I}_{\left[ 1,+\infty \right)}}\left( x \right)$, I found the following limit: $$\underset{x\to +\infty }{\mathop{\lim }}\,{{\left( 1-\exp \left( \frac{y+{{\left( \ln \left( x \right) \right)}^{2}}}{y-\ln \left( x \right)} \right) \right)}^{x}}.$$
The correct answer is $\exp(-\exp(-y))$.
How to proceed to find this result?
Thanks.
Let $g\left( x \right)={{\left[ 1-\exp \left( \frac{y+{{\ln }^{2}}\left( x \right)}{y-\ln \left( x \right)} \right) \right]}^{x}}$.
$\underset{x\to +\infty }{\mathop{\lim }}\,g\left( x \right)=\underset{x\to +\infty }{\mathop{\lim }}\,{{\left[ 1-\exp \left( \frac{y+{{\ln }^{2}}\left( x \right)}{y-\ln \left( x \right)} \right) \right]}^{x}}$ $=\underset{x\to +\infty }{\mathop{\lim }}\,{{\left[ 1-\exp \left( \frac{y-{{y}^{2}}+{{y}^{2}}-2y\ln \left( x \right)+2y\ln \left( x \right)+{{\ln }^{2}}\left( x \right)}{y-\ln \left( x \right)} \right) \right]}^{x}}$ $=\underset{x\to +\infty }{\mathop{\lim }}\,{{\left[ 1-\exp \left( \frac{y-{{y}^{2}}+2y\ln \left( x \right)+{{y}^{2}}-2y\ln \left( x \right)+{{\ln }^{2}}\left( x \right)}{y-\ln \left( x \right)} \right) \right]}^{x}}$ $=\underset{x\to +\infty }{\mathop{\lim }}\,{{\left[ 1-\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)+{{\left( y-\ln \left( x \right) \right)}^{2}}}{y-\ln \left( x \right)} \right) \right]}^{x}}$ $=\underset{x\to +\infty }{\mathop{\lim }}\,{{\left[ 1-\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\exp \left( \frac{{{\left( y-\ln \left( x \right) \right)}^{2}}}{y-\ln \left( x \right)} \right) \right]}^{x}}$ $=\underset{x\to +\infty }{\mathop{\lim }}\,{{\left[ 1-\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\exp \left( y-\ln \left( x \right) \right) \right]}^{x}}$ $=\underset{x\to +\infty }{\mathop{\lim }}\,{{\left[ 1-\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\exp \left( y \right)\exp \left( \ln \left( {{x}^{-1}} \right) \right) \right]}^{x}}$ $=\underset{x\to +\infty }{\mathop{\lim }}\,{{\left[ 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right]}^{x}}$
Let $h\left( x \right)={{\left[ 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right]}^{x}}$ $\ln \left( h\left( x \right) \right)=x\ln \left( 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right)$
$\underset{x\to +\infty }{\mathop{\lim }}\,\ln \left( h\left( x \right) \right)=\underset{x\to +\infty }{\mathop{\lim }}\,x\ln \left( 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right)$
$=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\ln \left( 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right)}{\frac{1}{x}}$
$\underset{\text{L }\!\!\acute{\ }\!\!\text{ H}}{\overset{\frac{0}{0}}{\mathop =}}\,\frac{\frac{-\exp \left( y \right)\left[ \exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{d}{dy}\left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x}+\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\left( -\frac{1}{{{x}^{2}}} \right) \right]}{\left( 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right)}}{-\frac{1}{{{x}^{2}}}}$$=\frac{\frac{-\exp \left( y \right)\left[ \exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\left( \frac{\frac{2y}{x}\left( y-\ln \left( x \right) \right)-y\left( 1-y+2\ln \left( x \right) \right)\left( -\frac{1}{x} \right)}{{{\left( y-\ln \left( x \right) \right)}^{2}}} \right)\frac{1}{x}-\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{{{x}^{2}}} \right]}{\left( 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right)}}{-\frac{1}{{{x}^{2}}}}$
$=\frac{\exp \left( y \right)\left[ \exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\left( \frac{2y\left( y-\ln \left( x \right) \right)+y\left( 1-y+2\ln \left( x \right) \right)}{{{\left( y-\ln \left( x \right) \right)}^{2}}} \right)-\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right) \right]}{\left( 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right)}$ $=\frac{\exp \left( y \right)\left[ \exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\left( \frac{2{{y}^{2}}-2y\ln \left( x \right)+y-{{y}^{2}}+2y\ln \left( x \right)}{{{\left( y-\ln \left( x \right) \right)}^{2}}} \right)-\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right) \right]}{\left( 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right)}$$=\frac{\exp \left( y \right)\left[ \exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\left( \frac{{{y}^{2}}+y}{{{\left( y-\ln \left( x \right) \right)}^{2}}} \right)-\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right) \right]}{\left( 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right)}$ $=\frac{\exp \left( y \right)\left[ \exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\left( \frac{{{y}^{2}}+y}{{{\left( y-\ln \left( x \right) \right)}^{2}}}-1 \right) \right]}{\left( 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right)}$
How $\underset{x\to +\infty }{\mathop{\lim }}\,\frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)}\underset{\text{L }\!\!'\!\!\text{ H}}{\overset{\frac{+\infty }{+\infty }}{\mathop{=}}}\,y\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\frac{2}{x}}{-\frac{1}{x}}=-2y$
$\underset{x\to +\infty }{\mathop{\lim }}\,g\left( x \right)=\exp \left( \underset{x\to +\infty }{\mathop{\lim }}\,\ln \left( h\left( x \right) \right) \right)$ $=\exp \left( \underset{x\to +\infty }{\mathop{\lim }}\,\frac{\exp \left( y \right)\left[ \exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\left( \frac{{{y}^{2}}+y}{{{\left( y-\ln \left( x \right) \right)}^{2}}}-1 \right) \right]}{\left( 1-\exp \left( y \right)\exp \left( \frac{y\left( 1-y+2\ln \left( x \right) \right)}{y-\ln \left( x \right)} \right)\frac{1}{x} \right)} \right)$ $=\exp \left( \frac{\exp \left( y \right)\left[ \exp \left( -2y \right)\left( 0-1 \right) \right]}{\left( 1-\exp \left( y \right)\exp \left( -2y \right)\cdot 0 \right)} \right)=\exp \left( -\exp \left( -y \right) \right)={{e}^{-{{e}^{-y}}}}$.
Thanks a lot.