We have a function $f(x)$ that is continuous for all $x\geqslant 0$. We were able to calculate this function as $$ f(x) = \frac{1}{x^n}\left[\Gamma(n) - \Gamma(n + 1, x)\right], $$ where $n \in \mathbb{N}^+$. Our goal is to analyze the monotonicity of this function but the problem is that I would like, if possible, to first show that $$f(x) =\begin{cases} 0 & x = 0, \\ \frac{1}{x^n}\left[\Gamma(n) - \Gamma(n + 1, x)\right] & x > 0. \end{cases}$$
If I'm not mistaken, to prove the second formulation holds, I need to show that $$\lim_{x \rightarrow 0^+} f(x) = 0.$$
But so far I have not been successful since we would get $\frac{1}{0}$. One thing I could think of was to write $$f(x) = \frac{1}{x^n}\left[\Gamma(n) - n! e^{-x} \sum_{k = 0}^{n} \frac{x^k}{k!} \right]$$ since $n$ is a positive integer but I couldn't progress further.
I also tried to expand it around $x = 0$ as a series but it didn't work. So is there a way to reformulate $f(x)$ to check if the above limit holds and later on, analyze the monotonicity of $f(x)$?
Edit
I have made a mistake. The function is $$f(x) =\begin{cases} 0 & x = 0, \\ \frac{1}{x^n}\left[\Gamma(n + 1) - \Gamma(n + 1, x)\right] & x > 0. \end{cases}$$
I do not know how much this could help you.
Around $x=0$, we have $$\Gamma(n+1,x)=\Gamma(n+1)+x^n \sum_{k=1}^\infty (-1)^k\frac{ x^k}{(k-1)!\,\, (k+n)}$$ So, if $$f(x) = \frac{1}{x^n}\left[\Gamma(n) - \Gamma(n + 1, x)\right]$$ then $$f(x)=-\frac{(n-1) \Gamma (n)} {x^n}+ \sum_{k=1}^\infty (-1)^{k+1}\frac{ x^k}{(k-1)!\,\, (k+n)}$$
So, it does not seem possible to have $\lim_{x \rightarrow 0^+} f(x) = 0$ (except if $n=1$)