Can we identify the pole of the Gamma function with the limit of the harmonic numbers? The expansion of the gamma function around $x=0$ is
$$\Gamma(x)=\frac{1}{x}-\gamma+O(x).$$
The Euler constant $\gamma$ is defined by $$\gamma=\lim_{n\to\infty}(H_n-\log n)$$ where $H_n$ is the $n$th harmonic number. On the other hand, $\gamma$ can also be defined as $$\gamma=\lim_{x\to 0}\left(\frac{1}{x}-\Gamma(x)\right).$$
However, $\lim_{n\to\infty} \log n$ and $\lim_{x\to 0} 1/x$ can be sort of identified in the following way:
$$\lim_{x\to 0^+} \frac{1}{x} = \lim_{x\to 0^+}\lim_{\delta\to 0}\int_\delta^1 t^{-1+x} dx$$
Now I will be sketchy and interchange the limits.
$$\lim_{x\to 0^+} \frac{1}{x} = \lim_{\delta\to 0}\lim_{x\to 0^+}\int_\delta^1 t^{-1+x} dt = \lim_{\delta\to 0}\int_\delta^1 t^{-1} dt
=-\lim_{\delta\to 0}\log\delta=\lim_{n\to\infty} \log n .$$
Then comparing to the two definitions of $\gamma$ above, it seems like one could again "identify" the pole of $\Gamma(x)$ with the limit of the harmonic numbers.
We can do a similar procedure to try to establish that.
$$\lim_{x\to 0^+} \Gamma(x)=\lim_{x\to 0^+} \lim_{\delta\to 0} \int_\delta^\infty t^{x-1} e^{-t} dt$$
Again being sketchy with the limits:
$$\lim_{x\to 0^+} \Gamma(x)=\lim_{\delta\to 0}  \int_\delta^\infty t^{-1} e^{-t} dt = \lim_{\delta\to 0}\Gamma(0,\delta)$$
Here $\Gamma(s,z)$ is the upper incomplete gamma function, which for $s=0$ has the following expansion around $z=0$.
$$\Gamma(0,z)=-\gamma-\log z+O(z)$$
Comparing this to the definition of $\gamma$ we obtain
$$\lim_{x\to 0^+} \Gamma(x) = -\lim_{n\to\infty}H_n.$$
My questions are: How legitimate is this? Is this a well-known statement? (I haven't seen it anywhere.) If it is legitimate, what is its significance?

What is true is that the two equations
$$ H_n = \log(n) + \gamma + \frac1{2n} + \cdots \tag{1}$$
and
$$ \Gamma(1/n) = n - \gamma + \frac1n + \cdots \tag{2} $$
together imply that
$$ \lim_{n\to\infty} \Gamma(1/n) + H_n - n - \log(n) = 0.
\tag{3} $$
