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?

$$
\big(1-x^{1/2x}\big)\cdot\frac{x}{2\ln{x}}=-\frac{1}{4}\frac{\mathrm{e}^{\ln x/2x}-1}
{\frac{\ln x}{2x}}=-\frac{1}{4}\frac{\mathrm{e}^t-1}{t},
$$
where $\,t=\dfrac{\ln x}{2x}$. But as $\,\lim_{x\to\infty}\dfrac{\ln x}{2x}=0,$ then we have
$$
\lim_{x\to\infty}\big(1-x^{1/2x}\big)\cdot\frac{x}{2\ln{x}}=\lim_{t\to 0}-\frac{1}{4}\frac{\mathrm{e}^t-1}{t}=-\frac{1}{4},
$$
as
$$
\lim_{t\to 0}\frac{\mathrm{e}^t-1}{t}=\left.\frac{d}{dt}\right|_{t=0}\mathrm{e}^t
=1.
$$
