Pointwise Convergence of reciprocal sequence of Pointwise Convergence Sequence. Assume that $f_n\rightarrow f$ pointwise on a set $E\subseteq \mathbb{R}$. Further assume that $f(x) \neq 0$ for all $x\in E$. Can we prove that $1/f_n \rightarrow 1/f$ pointwise on $E$? If not, then what other conditions will make this happen?
My proof:
For each $x\in E$, we can find an index $N_1$ such that
$$n\geq N_1 \text{ implies } |f_n(x)-f(x)| < |f(x)|/2$$
which further implies that (reverse triangle inequality) $|f_n(x)| > |f(x)|/2$.
Next, for any $\epsilon >0$, we can find an index $N_2$ such that
$$n\geq N_2 \text{ implies } |f_n(x) -f(x)| < \frac{\epsilon |f(x)|^2}{2} $$
Finally, for $n \geq \max \{N_1, N_2\}$ and using the above two inequalities,
$$ |(1/f_n)(x) -1/f(x)| \leq  \frac{|f_n(x) -f(x)|}{|f_n(x)f(x)|}< \epsilon $$
This is my proof. Let me know if you find any mistake or any other better idea. Thanks,
 A: It’s true of course! The proof should consist of repeating the proof of the analogous result for sequences. That is, assume $a_n \to a$, and $a\ne 0$. Then $1/a_n\to 1/a$. In proving this, you need to use that $a\ne 0$ to get a lower bound on $|a_n|$ for large values of $n$.
A: First we assume that $f_n(x)\neq 0$ for all $n\in\mathbb{N}$ and for all $x\in E$ .
We want to show that for each $x\in E$ :
$\forall\epsilon>0,\exists N\in\mathbb{N} : n\geq N\implies\left\lvert \frac{1}{f_n(x)} - \frac{1}{f(x)}\right\rvert<\epsilon$ where $f(x)\neq 0$
Consider $x\in E$, and let $\epsilon>0$. We have $\left\lvert \frac{1}{f_n(x)} - \frac{1}{f(x)}\right\rvert = \left\lvert \frac{f(x) - f_n(x)}{f_n(x)f(x)}\right\rvert=\frac{\lvert f(x) - f_n(x)\rvert}{\lvert f_n(x)f(x)\rvert}$
but there exists $N_1\in\mathbb{N}$ such that $n\geq N_1\implies \left\lvert  \lvert f_n(x)\rvert - \lvert f(x)\rvert\right\rvert\leq\lvert f_n(x) - f(x)\rvert < \frac{\lvert f(x)\rvert}{2}\implies\lvert f_n(x)\rvert\geq\frac{\lvert f(x)\rvert}{2} $ (the last implication follows by a caracterization of the absolute value)
and for $\epsilon_1 = \frac{\epsilon\lvert f(x)\rvert^{2}}{2}$ there exists $N_2\in\mathbb{N}$ such that $n\geq N_2\implies \lvert f_n(x) - f(x)\rvert <\epsilon_1$
Then by taking $N=max\{N_1, N_2\}$ we have
$n\geq N\implies\left\lvert \frac{1}{f_n(x)} - \frac{1}{f(x)}\right\rvert = \left\lvert \frac{f(x) - f_n(x)}{f_n(x)f(x)}\right\rvert=\frac{\lvert f(x) - f_n(x)\rvert}{\lvert f_n(x)f(x)\rvert}\leq\frac{2\lvert f(x)-f_n(x)\rvert}{\lvert f(x)\rvert^{2}} < \frac{2\epsilon_1}{\lvert f(x)\rvert^{2}} = \epsilon$
Since this holds for an arbitrary $x\in E$ we conclude that
$\forall x\in E,\forall\epsilon>0,\exists N\in\mathbb{N} : n\geq N\implies\left\lvert \frac{1}{f_n(x)} - \frac{1}{f(x)}\right\rvert<\epsilon$
So your proof is correct yes !
