Limits Again. Ekdum arginal

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to \infty} (x\ln 2)^{2^{-x}} \\ & = \lim_{x \to \infty} x^{\frac 1{2^x}} (\ln 2)^{\frac 1{2^x}} & \small \color{#3D99F6} \text{Let }u = 2^x \implies \ln u = x \ln 2 \\ & = \lim_{u \to \infty} \left(\frac {\ln u}{\ln 2}\right)^\frac 1u (\ln 2)^\frac 1u \\ & = \lim_{u \to \infty} (\ln u)^\frac 1u \\ & = \lim_{u \to \infty} \exp \left(\frac {\ln(\ln u)}u\right) & \small \color{#3D99F6} \text{where }\exp(x) = e^x \\ & = \lim_{u \to \infty} \left(1 + \frac {\ln(\ln u)}u + \frac 1{2!} \left(\frac {\ln(\ln u)}u\right)^2 + \cdots\right) & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \boxed{1} & \small \color{#3D99F6} \text{Since }\lim_{u \to \infty} \frac {\ln(\ln u)}u = 0 \text{ (see note)} \end{aligned}$

Note:

$\begin{aligned} L_u & = \lim_{u \to \infty} \frac {\ln(\ln u)}u & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \lim_{u \to \infty} \frac {\frac 1{u\ln u}}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }u \\ & = 0 \end{aligned}$

Note that the substitution only works because when you will differentiate 2^x in both numerator and denominator 2^x(ln2) will cancel out . So it is the same as writing it as another variable t . Also from the third step I should write it as limit of t tends to infinity instead of x after the substitution. Apologies for that.