Convergence of metrics Let $U\subset\mathbb{R}^2$ be an open neighborhood of the origin and let $f_n:U\rightarrow \mathbb{R}_{>0}$, $n\in \mathbb{N}$ be a sequence of differentiable functions which uniformly converges on $U$ to an integrable function $f:U\rightarrow \mathbb{R}_{>0}$. 
Fix two point $p_1,p_2\in U$ and call $\Gamma= \Gamma_{p_1}^{p_2}$ the set of differentiable paths $\gamma:I\rightarrow U$, $\gamma(t)=(\gamma_1(t),\gamma_2(t))$, such that $\gamma(0)=p_1$, $\gamma(1)=p_2$ and $\gamma(I)\subset U$.
For every $\gamma\in \Gamma$ I'm quite sure it's true: 
$$\lim_{n\rightarrow \infty}\int_0^1\sqrt{f_n(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt=\int_0^1\sqrt{f(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt,$$ 
since the functions $f_n$ converge uniformly.
I'm not sure if it's also true 
$$\lim_{n\rightarrow \infty}\inf_{\gamma\in\Gamma}\int_0^1\sqrt{f_n(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt
= \inf_{\gamma\in\Gamma}\int_0^1\sqrt{f(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt
$$
Is this equality true or is it only verified under additional hypothesis?
 A: This equality is true. For each $\epsilon>0$ we can choose a path $\gamma$ between $p_1$ and $p_2$ such that $l_g(\gamma)< d_g(p_1,p_2)+\epsilon$.  Since $f_n$ converges to $f$ uniformly, we can pick an $N>0$ such that for each $n>N$ we have also $l_{g_n}(\gamma)<l_g(\gamma)+\epsilon$.  Therefore $d_{g_n}(p_1,p_2)\leq l_{g_n}(\gamma)<d_g(p_1,p_2)+2\epsilon$.  Since $\epsilon$ is arbitary, we obtain $\lim_{n\to\infty} d_{g_n}(p_1,p_2)\leq g(p_1,p_2)$.
Since the path $\gamma$ is compact, $\frac{f_n}{f}$ converges to $1$ uniformly along $\gamma$.  Therefore $l_{g_n}(\gamma)$ converges to $l_g(\gamma)$, proving the opposite inequality.
In more detail, given points $p,q$, the sequence $d_{g_n}(p,q)$ converges to some limit $L>0$.  For small $\epsilon>0$ negligible compared to $L$ choose $n$ large enough so that $d_{g_n}(p,q)>L-\epsilon$. Then for each curve $\gamma$ between $p$ and $q$ we have $l_{g_n}(\gamma)>L-\epsilon$. In particular if $\gamma$ is a minimizing curve for $g$ we argue as above to get that $L=d_g(p,q)$. If $\gamma$ is not minimizing choose a sufficiently good approximation and argue as before.
A: I want to add more in opposite direction-proof of Mikhail Katz's answer
(1) If $d_g(p,q) >\delta+d_n(p,q)$ for all $n$, there is $N_1$ s.t.
$n>N_1\Rightarrow |f_n-f|<\varepsilon_1$
If $c_n$ is a geodesic from $p$ to $q$ wrt $g_n$ then
\begin{align*} d_g (p,q) &\leq \int g(c_n',c_n')^\frac{1}{2} \\&\leq
\int \sqrt{ g_n(c_n',c_n') +\varepsilon_1 h(c_n',c_n')} \\&\leq
d_n(p,q) +\sqrt{\varepsilon_1} {\rm length}_h\ (c_n)
\end{align*}
 where $h=dx^2+dy^2$
Hence $$ \delta <\sqrt{\varepsilon_1} {\rm length}_h\ (c_n) $$
That is $\lim_n\ {\rm length}_h \ (c_n) =\infty$ Then $$ d(p,q) >
d_n(p,q)=\int_0^{L_n} \sqrt{f_n(t)} dt
$$ where $c_n(t)$ has unit speed wrt $h$ and ${\rm length}_h \
c_n=L_n$
Hence if all $c_n$ is in $B^{h}(p,R)$ for some $R>0$, we have a
point $p_n$ in $c_n$ s.t. $f_n(p_n)\rightarrow 0$ This contradicts
uniform
convergence
So there is $p_n$ in $c_n$ s.t. $d_h(p,p_n)\rightarrow \infty\ \ast$
(2) Consider a sequence of ball : For $q_n\in
B_{r_n}^{g_n} (p)$ there is a unique geodesic from $p$ to $q_n$ We
have a claim that $\bigcap_n B_{r_n}^{g_n} (p)\supseteq B_r^h(p) $ :
If we choose $q\in B_r^h(p) $, then $\ast$ is impossible
Proof of claim : If not there is $q_n$ s.t. $d_h(p,q_n)\rightarrow
0$ and there are two geodesics from $p$ to $q_n$ wrt $g_n$ This
means that ${\rm Rm}\ (g_n,s_n)\rightarrow \infty $ for some point
$s_n$ around $p$ Since curvature of $g$ around $p$ is bounded above
so this is impossible.
