Why this limit is $-\frac{1}{4}$? Find
$$\lim_{x\rightarrow\infty} \bigg(1-x^{1/2x}\bigg)\cdot\frac{x}{2\ln{x}}$$
I tried this method:
$$\begin{align}
\bigg(1-x^{1/{2x}}\bigg)\frac{x}{2\ln{x}} & = \frac{x}{2}\frac{1-x^{1/{2x}}}{\ln{x}}\frac{1+x^{1/{2x}}}{1+x^{1/{2x}}} \\
& = \frac{1}{2} \frac{x}{\ln{x}} \frac{1-\sqrt[x]{x}}{1+\sqrt{\sqrt[x]{x}}} \\
& = \frac{1}{2 \cfrac{\ln{x}}{x} \cfrac{1+\sqrt{\sqrt[x]{x}}}{1-\sqrt[x]{x}}} \\
\end{align}$$
I know that whenever $x \to \infty$, so does $\frac{\ln{x}}{{x}}\to{0}$. Likewise, $\sqrt[x]{x}\to{1}$.
If I apply L'Hopital to last altered equation, the fraction will blow off to infinity. My approach is not the best, can you help me?

Note that $e^x-1\sim x$ as $x\to0$,
$$(1-x^{1/2x})\frac{x}{2\ln x}\sim(1-e^{(\ln x)/2x})\frac{x}{2\ln x}\sim-\frac{\ln x}{2x}\frac{x}{2\ln x}\sim-\frac14$$

To understand how to move the second to the third, apply L'Hospital rule on the first factor
$$1-e^{(\ln x)/2x}=\frac{1-e^{(\ln x)/2x}}{(\ln x)/2x}\frac{\ln x}{2x}\sim-\frac{e^{(\ln x)/2x}\Big((\ln x)/2x\Big)'}{\Big((\ln x)/2x\Big)'}\frac{\ln x}{2x}\sim-e^{(\ln x)/2x}\frac{\ln x}{2x}$$
