If $f_n(x)$ converge to $f(x)$ uniformly on a domain E and if $f_n$ $f$ never vanish on E ($f_n(x)\neq0$ and $f(x)\neq0$) $\forall x\in E$ and n in natural numbers
then does it follow that the functions $1/f_n$ converge uniformly to $1/f$ on E?
I am not sure how to do this one.
I believe you need some stronger bounded-above-zero conditions.
Take $f_n(x) = 1/x + 1/n$ over $E = (0,\infty)$ and $f(x) = 1/x$. Note that $$|f_n(x) - f(x)| = 1/n < \varepsilon$$ for any $\varepsilon$ for sufficiently large $n$, so we have uniform convergence. Now, $$\left| \frac{1}{f_n(x)} - \frac{1}{f(x)} \right| = \left| \frac{xn}{n+x} - x\right| = \frac{x^2}{n+x}.$$ For fixed $n$, the supremum of the last quantity over $E$ is $+\infty$, so we do not have uniform convergence. We do have pointwise convergence, however!