Show that $\frac{1}{f_{n}}$ converges uniformly to $\frac{1}{f}$ given the following Please point out any flaws in my arguments
Let $f_{n}$ be a sequence of functions on $[a,b]$ s.t. $f_{n}(x) \neq 0$ $\forall x \in [a,b], n \in \mathbb{N}$ and let $f : [a,b] \to \mathbb{R}$ and $\exists L > 0$ s.t. $f(x) \geq L$ $\forall x \in [a,b]$. Now suppose $f_{n}$ converges uniformly to $f$ on $[a,b]$.
I want to show that $\frac{1}{f_{n}}$ converges uniformly to $\frac{1}{f}$ on $[a,b]$ i.e. given $\varepsilon > 0$, find $N \in \mathbb{N}$ s.t. $\forall n \geq N, x \in [a,b]$,
$$|\frac{1}{f_{n}(x)} - \frac{1}{f(x)}| = |\frac{f(x) - f_{n}(x)}{f_{n}(x)f(x)}| < \varepsilon.$$
Since $f_{n}$ converges uniformly to $f$ on $[a,b]$, $\exists M \in \mathbb{N}$ s.t. $\forall m \geq M, x \in [a,b]$,
$$|f_{n}(x) - f{x}| < \frac{\varepsilon L}{r(x)}$$
where $r(x) \in \mathbb{N}$ is any integer s.t.
$$0 < \frac{1}{r(x)} < |f_{n}(x)|$$
$\forall x \in [a,b]$ (since $\forall x \in [a,b], f_{n}(x) \neq 0 \implies |f_{n}(x)| > 0$, it follows from the archimedean property that r(x) exists).
Then for $n \geq M, x \in [a,b]$,
$$|\frac{1}{f_{n}(x)} - \frac{1}{f(x)}| = |\frac{f(x) - f_{n}(x)}{f_{n}(x)f(x)}| = \frac{|f(x) - f_{n}(x)|}{|f_{n}(x)||f(x)|} < \frac{r(x) \cdot \varepsilon L}{r(x) \cdot L} = \varepsilon.$$
 A: I don't think what you did is correct. Because the way you have used $r(x)$ (it should have been $r(n)$ as it depends only on $n$ and not on $x$), it is a function of $n$. As $n$ varies $r(x)$ (or $r(n)$) also varies. Hence $\frac{\epsilon L}{r(x)}$ is not a fixed quantity. Hence direct application of definition of uniform convergence is not allowed.
Instead we could have shown that there exists an $N \in \mathbb{N}$ and $r>0 \in \mathbb{R}$, such that $|f_n(x)| > r$, for all $n\geq N$ and for all $x$. In other words the set $\{|f_n(x)| \quad | n \geq N, x\in [a,b]\}$ is bounded below by $r$. If this is not true, then given any $\epsilon > 0$, we can choose $x_{n_k}$ such that $f_{n_k}(x_{n_k}) < \frac{\epsilon}{2}$ for some subsequence $n_k$ of natural numbers. Now, by uniform convergence condition, there exists an $M$ such that for all $n\geq M$, $|f_n(x)-f(x)| < \frac{\epsilon}{2}$ for all $x$. Choose $n_k > M$, then we have
$$|f(x_{n_k})| =|f(x_{n_k})-f_{n_k}(x_{n_k})+ f_{n_k}(x_{n_k})| \leq |f(x_{n_k}) - f_{n_k}(x_{n_k})| + |f_{n_k}(x_{n_k})| < \epsilon.$$
In other words, for each $\epsilon$ we can find an element $x$ in $[a,b]$ such that $f(x) < \epsilon$. Which is not true by hypothesis. Hence our assertion.
Now fix that $N$ and $r$. There exists $N_1$ such that for all $n\geq N_1$, it is true that
$$ |f(x)- f_n(x)| < \frac{\epsilon L r},$$
for all $x\in [a,b]$. Choose $N_2 = max \{N, N_1\}$. Then for all $n\geq N_2$, we have
$$|\frac{1}{f_n(x)}-\frac{1}{f(x)}| < \epsilon.$$
