How can we show that $3\gamma=\lim_{x\to \infty}\left(xe^{1\over x}-e^{-{1\over x}}\Gamma\left({1\over x}\right)\right)?$

Question

How can we show that

$$3\gamma=\lim_{x\to \infty}\left(xe^{1\over x}-e^{-{1\over x}}\Gamma\left({1\over x}\right)\right)?\tag1$$

Where $\gamma=0.577...$ is Euler-Masheronic Constant

It is all known that

$$\gamma=\lim_{x\to \infty}\left(x-\Gamma\left(1\over x\right)\right)\tag2$$

Answer 1

Consider $x=1/t$ to get

$$L=\lim_{t\to0^+}\frac1te^t-\Gamma(t)e^{-t}$$

Now note that

$$\Gamma(t)=\frac{\Gamma(t+1)}t$$

Thus,

\begin{align}L&=\lim_{t\to0^+}\frac{e^t-1}t-\frac{\Gamma(t+1)e^{-t}-1}t\\&=1-\frac d{dt}\Gamma(t+1)e^{-t}\bigg|_{t=0}\\&=2-\Gamma'(1)\\&=2+\gamma\end{align}

Which is the correct limit (you can check it numerically)