Let $f_n:M \to \mathbb{R}$ be bounded by $f_n(x) \geq a > 0$ for all $x \in M, n \in \mathbb{N}$ and uniformly convergent in $M$ towards $f$.
How can one prove that $\frac{1}{f_n}$ converges uniformly towards $\frac{1}{f}$ in $M$ as well?
I know that the definition of uniform convergence states that a sequence of functions $f_n(x)$ , which converges to $f(x)$, converges uniformly to $f(x)$ for $x\in M$ if for all $\epsilon>0$, there exist a number $N\in \mathbb{N}$, such that for all $x\in M$ and all $n>N$
$$|f_n(x)-f(x)|< \epsilon$$
Let $M=\{x\,|\,\delta \le x\le 1\}$, for $\delta>0$ and let $\epsilon>0$ be given. Then
$$\left|f_n(x)-0\right|=\frac{1}{nx}\le \frac{1}{n\delta }<\epsilon$$
for all $x\in M$ and for all $n>\frac{1}{\delta \epsilon}$.
So, $f_n(x)$ converges uniformly to $0$ for $x\in [\delta,1]$ for $\delta >0$.
Is this correct or is there an alternative way to prove this?
If $f_n\to f$ uniformly, then $1/f$ is not the $0$ function, nor is $f_n$ uniformly bounded by below. However, you can just use the definition to prove the statement, yes.
$$ \left|\frac{1}{f_n(x)}-\frac{1}{f(x)}\right|=\left|\frac{f_n(x)-f(x)}{f_n(x)f(x)}\right|\leq \frac{1}{2a} |f_n(x)-f(x)|, $$ and now, you can apply uniform convergence of $f_n$.