Limit of the gamma function I have to prove that $$\lim_{x\to 0^{+}}\Gamma(x)=\lim_{x\to+\infty}\Gamma(x)=+\infty$$
where $\Gamma$ is the gamma function.
As for the $x\to\infty$ I would like to show that $\Gamma$ is increasing on interval $(a,\infty)$ for some $a>0$. Increasing function has a limit and $$\lim_{x\to+\infty}\Gamma(x)=\lim_{n\to+\infty}\Gamma(n)=\lim_{n\to+\infty}(n-1)!=\infty$$
But I don't know how to prove the function is increasing.
As for the $x\to 0$ my idea is
$$\lim_{x\to 0^{+}}\Gamma(x)=\lim_{x\to 0^{+}}\int_{0}^{\infty}e^{-t}t^{x-1}dt=\int_{0}^{\infty}\lim_{x\to 0^{+}}e^{-t}t^{x-1}dt=$$$$=\int_0^{\infty}e^{-t}t^{-1}dt\ge\int_0^1e^{-t}t^{-1}dt\ge e^{-1}\int_0^1t^{-1}dt=\infty$$
But I don't know whether I can change the order of limit and integral (second equality). Maybe if I showed that $\Gamma$ is decreasing on interval $(0,a)$ for some $a>0$ I could replace the limit $x\to 0$ with $n\to\infty$ and use some Lebesgue theorem.
 A: The $\Gamma$ function is log-convex, hence convex, by the Bohr-Mollerup theorem (we may take that theorem as the definition of the $\Gamma$ function, too, since it gives the Euler product as a by-product).
The $\Gamma$ function is positive on $(0,1)$ as the integral of a positive function, hence the functional relation $\Gamma(x+1)=x\cdot\Gamma(x)$ gives that $\Gamma(x)>0$ for any $x>0$. $\Gamma(x)$ is increasing over $(2,+\infty)$ because: 
$$\frac{d}{dx}\log\Gamma(x) = \psi(x) = -\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{x-1+n}\right)$$
and the RHS, given $x>2$, is positive:
$$ -\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{x-1+n}\right)\geq -\frac{2}{3}+\sum_{n\geq 1}\frac{1}{n(n+1)} = \frac{1}{3}.$$
Moreover, since $\Gamma(1)=1$ and we have continuity in a neighbourhood of $1$ (as a consequence of convexity),
$$ \lim_{x\to 0^+}\Gamma(x) = \lim_{x\to 0^+}\frac{\Gamma(x+1)}{x} = \lim_{x\to 0^+}\frac{1}{x} = +\infty.$$
A: You can prove using integration by parts that for all $x>0$
$$\Gamma(x+1)=x\Gamma(x)$$
since $\Gamma$ is continuous, then
$$\lim_{x\to 0^+}x\Gamma(x)=\lim_{x\to 0^+}\Gamma(x+1)=\Gamma(1)=1.$$ Thus
${\displaystyle \Gamma(x)\underset{0}{\sim}\frac{1}{x}}$ which means $\Gamma(x)$ and $\displaystyle\frac{1}{x}$ have the same limit at $0$.
You can also interchange limit and integral using monotone convergence theorem as suggested by
Daniel Fischer, or use Fatou's lemma.
