Limit of Subexponential Function During my cryptology course, the professor mentioned to us that we could achieve better results, as the problem says, if we did the following.  The issue is that he never fully explained to us the why behind it, so I am currently giving this problem a go but can't seem to grasp how to proceed.
I am giving part b a go first because I normally feel comfortable with limits, but I already tried simplifying and I attempted using L'Hopital's Rule, but both methods got really messy.  I also attempted using the Limit Definition of Convergence but that still did not get me anywhere.
I feel like I am missing something, so I would really appreciate any input.
Note: 
$L(X)= e^{\sqrt{lnX lnlnX}}$

 A: As pointed out by @Clement C., the limit in exercise (b) is 
$$
X= \lim_{N->\infty}\frac{\log\left[L(\sqrt{N})\right]}{\log\left[ (L(N)^{1/\sqrt{2}}\right]} =\sqrt{2}
\lim_{N->\infty}\frac{\log\left[L(\sqrt{N})\right]}{\log\left[ (L(N)\right]} 
$$
Let $u = \log\log(N)$; when $n\to\infty$ we have $u\to\infty$.  Also, 
$$L(N) = \sqrt{u \log N} = \sqrt{u e^u} 
$$
and
$$
L(\sqrt{N}) =  \sqrt{\log \sqrt{N} \log\log\sqrt{N}} = 
\sqrt{\log \left( e^{\frac12 e^u}  \right)  \log \left(\frac12 \log N\right)}
= \sqrt{ \left( {\frac12 e^u}  \right)  \left( \log \frac12 +  \log\log N\right)} 
= \sqrt{ \left( {\frac12 e^u}  \right)  \left( \log \frac12 +   u\right)} \\ = \frac{1}{\sqrt{2}}
\sqrt{  {e^u}    \left( \log \frac12 + u\right)} 
$$
Then 
$$
X= \sqrt{2}
\lim_{u->\infty}\frac{\frac{1}{\sqrt{2}}
\sqrt{  {e^u}    \left( \log \frac12 + u\right)} }{\sqrt{u e^u} } \\ 
= \sqrt{2}\frac{1}{\sqrt{2}}
\lim_{u->\infty}\frac{\sqrt{u} } {
\sqrt{     u-\log{2}} } =\lim_{u->\infty}\left( 1 + \frac{\log 2}{2u} + O(u^{-2})\right) = 1
$$
where in the very last step we have used the fact that $\log 2$ is negligible compared to $u$ as $u$ goes to infinity.
