$\Gamma$ function is continuous on $]0,\infty [$ I would like to show that $\Gamma$ is continuous on $]0,\infty [$.
Let $x>0$,
$$|\Gamma(x+h)-\Gamma(x)|\leq\int_0^\infty t^{x-1}e^{-t}|t^h-1|dt=\underbrace{\int_0^1 t^{x-1}e^{-t}\underbrace{|t^h-1|}_{\leq |h|}dt}_{\underset{h\to 0}{\longrightarrow0 }}+\int_1^\infty t^{x-1}e^{-t}|t^h-1|dt$$
If $t> 1$, there is a $\delta>0$ such that $|t^h-1|\sim \ln(t)h\leq th$,
therefore, if $|h|<\delta$, 
$$\int_1^\infty t^{x-1}e^{-t}|t^h-1|dt\leq h\int_1^\infty t^{x}e^{-t}dt\leq |h|\Gamma(x+1)\underset{h\to 0}{\longrightarrow }0 .$$
We conclude that $$|\Gamma(x+h)-\Gamma(x)| \underset{h\to 0}{\longrightarrow }0$$
and thus $\Gamma$ is continuous on $]0,\infty [$.
Questions
1) Is this proof correct ?
2) I can't use theorem of dominated convergence. But If I were authorized to use it, would it be possible to use it here ? And if yes, how ?
Thanks,

1: Yes, your proof is correct:
2: For dominated convergence, you need a dominating function.  Consider the function
$$
g(x) = \max\{t^{x + \delta}e^t,t^{x - \delta}e^t\}
$$
Where we can choose any $\delta \in (0, x)$.  Once we know that this function is integrable, dominated convergence applies.
