$f_n(x): (0,1) \to (0,1)$ is a sequence of continuous functions that converges uniformly to $f(x): (0,1) \to (0,1)$. Is it true that $$ \sup_{x \in (0,1) }\frac{f_n(x)}{f_{n+1}(x)} \to 1 $$
I think pointwise convergence follows easily: since $\{f_n\}$ converges uniformly (but pointwise would suffice), then $|f_n(x)-f(x)| < \epsilon_n$ with $\epsilon_n \to 0$, so that: $$ \frac{f_n(x)}{f_{n+1}(x)} \leq \frac{f(x)+\epsilon_n}{f(x)-\epsilon_{n+1}} \to \frac{f(x)}{f(x)}=1 $$ $$ \frac{f_n(x)}{f_{n+1}(x)} \geq \frac{f(x)-\epsilon_n}{f(x)+\epsilon_{n+1}} \to \frac{f(x)}{f(x)}=1 $$
I am not really sure how to proceed to prove uniform convergence of such ratio (which may actually not hold). Indeed the previous approach does not work, as $f(x)-\epsilon_{n+1}$ may be equal to 0 for some $x \in (0,1)$ - unless we make the stronger assumption that $f(x) > c$ for all $x \in (0,1)$.
If you could provide any help or counter-examples that would be great!
Let $f_{n}(x)=x^{\frac{n+1}{n+2}}$ and $f(x)=x$, then clearly $0<f_n(x)<1$ and $0<f(x)<1$ for all $x \in (0,1)$.
Next, let's show $f_n(x)$ converges to $f(x)$ uniformly on $(0,1)$:
$\displaystyle h_n(x):= f_n(x) - f(x) = x^{\frac{n+1}{n+2}} - x = x \left( \frac{1}{\sqrt[n+2]{x}} - 1 \right)$.
$\displaystyle [h_n(x)]' = \frac{n+1}{n+2} \frac{1}{\sqrt[n+2]{x}} - 1 = 0 \implies \frac{n+1}{n+2} \frac{1}{\sqrt[n+2]{x}} = 1 \implies x = \left( \frac{n+1}{n+2}\right)^{n+2}$.
Thinking graphically, it can be easily verified that these points maximize $h_n(x)$ on $(0,1)$. Thus,
$\displaystyle \sup_{x \in (0,1)} |f_n(x) - f(x)| = \left( \frac{n+1}{n+2}\right)^{n+2} \frac{1}{n+1} = \left(1 - \frac{1}{n+2}\right)^{n+2} \frac{1}{n+1} \to \frac{1}{e}0 =0$ as $n \to \infty$.
Hence, $f_n(x)$ converges to $f(x)$ uniformly on $(0,1)$. However,
$\displaystyle \frac{f_{n}(x)}{f_{n+1}(x)} = \frac{x^{\frac{n+1}{n+2}}}{x^{\frac{n+2}{n+3}}} = \frac{1}{x^{1/(n+2)(n+3)}}$ gives us $\displaystyle \sup_{x \in (0,1)} \frac{f_{n}(x)}{f_{n+1}(x)} = \sup_{x \in (0,1)} \frac{1}{x^{1/(n+2)(n+3)}} = \infty$ for all $n$.