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?
 A: 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
$$1-e^{(\ln x)/2x}=\frac{1-e^{(\ln x)/2x}}{(\ln x)/2x}\frac{\ln x}{2x}\sim-\frac{e^{(\ln x)/2x}\Big((\ln x)/2x\Big)'}{\Big((\ln x)/2x\Big)'}\frac{\ln x}{2x}\sim-e^{(\ln x)/2x}\frac{\ln x}{2x}$$
A: $$
\big(1-x^{1/2x}\big)\cdot\frac{x}{2\ln{x}}=-\frac{1}{4}\frac{\mathrm{e}^{\ln x/2x}-1}
{\frac{\ln x}{2x}}=-\frac{1}{4}\frac{\mathrm{e}^t-1}{t},
$$
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
$$
\lim_{t\to 0}\frac{\mathrm{e}^t-1}{t}=\left.\frac{d}{dt}\right|_{t=0}\mathrm{e}^t
=1.
$$
A: Try with $t=1/x$, that changes the limit into
$$
\lim_{t\to0^+}-\left(1-\frac{1}{t^{t/2}}\right)\frac{1}{2t\ln t}=
\lim_{t\to0^+}\frac{1-t^{t/2}}{t^{t/2}}\frac{1}{2t\ln t}
$$
The denominator $t^{t/2}$ goes to $1$, because
$$
\lim_{t\to0^+}\frac{t}{2}\ln t=0
$$
so we're left with
$$
\lim_{t\to0^+}\frac{1-t^{t/2}}{2t\ln t}=
\lim_{t\to0^+}\frac{1-t^{t/2}}{2t\ln t}\frac{1+t^{t/2}}{1+t^{t/2}}=
\frac{1}{4}\lim_{t\to0^+}\frac{1-t^t}{t\ln t}=
\frac{1}{4}\lim_{t\to0^+}\frac{1-e^{t\ln t}}{t\ln t}
$$
The function $t\mapsto t\ln t$ is invertible (and assumes negative values) in an interval $(0,\varepsilon)$, so we can use the inverse and transform the last limit into a known limit:
$$
\frac{1}{4}\lim_{u\to0^{-}}\frac{1-e^u}{u}=-\frac{1}{4}
$$
A: $$\lim_{x\rightarrow\infty} \bigg(1-x^{\cfrac{1}{2x}}\bigg)\cdot\frac{x}{2\ln{x}}$$
Let $$f(x)=\bigg(1-x^{\cfrac{1}{2x}}\bigg)$$ 
and$$ g(x)=\frac{x}{2\ln{x}}$$
and $$H(x)=f(x)\cdot g(x)$$
as $$\lim{x\rightarrow\infty }$$ $g(x)\rightarrow \infty$ using L'Hospital rule 
and $f(x)\rightarrow 0$
so it is the  form of $0 \cdot \infty$
therefore covert it into  $\frac{\infty}{\infty}$
To do this take $ln$ on both side and proceed further 
and use L'Hospital rule . This will give $$\frac{-1}{4}$$
A: Let $y=g(x)=\frac{\ln x}{2x}$ so that $f(x)=(1-x^{1/2x})\frac{x}{2\ln x}$ can be rewritten as
$$f(x)=(1-x^{1/2x})\frac{x}{2\ln x}=\left(1-e^{(\ln x)/2x}\right)\frac{x}{2\ln x}=-\frac{1}{4}\frac{\operatorname{e}^{y}-1}{y}$$
Now you now that $\lim_{x\to\infty}g(x)=0$ applying l'Hopistal's rule for example $$\lim_{x\to\infty}\frac{\ln x}{2x}=\lim_{x\to\infty}\frac{1}{2x}=0$$
so you have
$$
\lim_{x\to\infty} f(x)=\lim_{y\to 0} -\frac{1}{4}\frac{\operatorname{e}^{y}-1}{y}=-\frac{1}{4}
$$
recalling the fundamental limit $\lim_{y\to 0} \frac{\operatorname{e}^{y}-1}{y}=1$.
