Consider the following ordinary differential equation. $$u''+f(u)=0, \tag1$$ where $''$ stands for second derivative, and $f\in C^1(-\infty,\infty)$, $f(0)=0,\,f'(0)=1$. Multiply both sides by $u'$ and integrate over $x$ from $a$ to $x$, $$\frac12u'(x)^2+F(u(x))=\frac12u'(a)^2+F(u(a))=\frac{c^2}2 \tag2$$ where $F(u)=\int_0^u f(t)dt$ and some positive constant $c$.
Suppose $c$ is very small. It is plausible that $F(u)=\frac12u^2+O(u^3)$ and $$\frac{c^2}2=\frac12u'(x)^2+F(u(x))\approx \frac12u'(x)^2+\frac12u^2$$ or as $u$ depends on both $c$ and $x$, $$u(c,x)\approx c\sin x, \tag3$$ and $$\lim_{c\rightarrow 0}\frac{u(c,x)}{c}=\sin x. \tag4$$
How does one rigorously justify this, perhaps in terms of asymptotic expansion and expand further in perturbative fashion?
Consider the Cauchy problem: $u(0)=0$, $u'(0)=c$ with $c\neq 0$. We write the solution as $u(t)=cw(t)$. Then $w(t)=w_c(t)$ verifies the ode: $$ w''(t)+\frac{1}{c} f(cw(t)) = 0, \ \ w(0)=0, w'(0)=1$$ The hypothesis on $f$ shows that $g(c,w)=f(cw)/c=\int_0^w f'(cs)\;ds$ is $C^1$ in $w$ and if we set $g(0,w)=w$ it is also continuous in $c$ in a neighborhood of $0$. The problem for $c=0$ is simply $w_0''+w_0=0, w_0(0)=0,w_0'(0)=1$ so $w_0(t)=\sin(t)$ and by continuity of solutions to ode's we have $w_c(t)-\sin(t)$ tends to zero uniformly on compact intervals as $c\rightarrow 0$.
If you assume $f$ is $C^r$, $r\geq 1$ then the function $g$ becomes $C^{r-1}$ in all variables so that the solution will be $C^{r-1}$ in parameters. So $w_c(t)$ is $C^1$ in both $c$ and $t$ [V.I.Arnold, Ordinary Diff Eqns, Chap 2, sec 7]. If $f$ is only $C^0$ (but differentiable at 0 with $f'(0)=1$) then I somehow suspect you still get continuity of solutions at $c=0$, but I wouldn't know (at least for now) how to prove this.
Looking in the $(u,v=u')$ plane the function $ H(u,v)=\frac12 v^2 + F(u)=\frac12 (u^2+v^2) + o(u^2)$ is a first integral of the flow. For $c$ small enough $$ K_c =\{ H(u,v) = \frac{c^2}{2} \} \cap \{u^2+v^2<2c^2\}$$ is compact connected, which implies that a solution starting in $K$ (our case) exists at all times. [see e.g. Arnold, chap 2, section 7, The extension thm]
To get more information, we go to the $w=u/c,w'=u'/c$ coordinates: We have $w''+g(w,c)=0$ and the function $G(c,w)=\int_0^{w} \frac{1}{c}f(ct)dt= \frac{1}{c^2} \int_0^{wc} f(u) du$ extends continuously to $G(0,w)=w^2/2$. The function $H(c,w,w')=\frac12 (w')^2 + G(c,w)$ is then a 1st integral for the flow. A level curve of $$\hat{K_c}=\{H(c,w,w')=1\} \cap \{w^2+(w')^2<3\}$$ is a circle for $c=0$ and it is regular (non-vanishing derivative/gradient) so using the implicit function theorem we see that for fixed $c$ small enough, $\hat{K}_c$ will be diffeomorphic to a circle as well.
Consider the solution curve $(w_c(t),w'_c(t))$ with $w_c(0)=0, w'_c(0)=1$. For $c=0$ we know that $w_0(T_0)=0$, $w'_0(T_0)=1$ with $T_0=2\pi$. Now, $$ (c,t) \rightarrow w_c(t) $$ is a smooth map and the $t$-derivative at $(c=0,T_0)$ is 1 (so non-zero). By the implicit function thm there is a $C^{r-1}$ map: $c\rightarrow T(c)$ (defined in a nghb of 0) so that $w_c(T(c))=0$. But then $w'_c(T(c))=1$ because of the first integral. The solution is thus periodic with a period that depends $C^{r-1}$ upon $c$ (small).