Proving that $\lim\limits_{x\to 0}\frac{f(x)}{g(x)}=L$ implies $\lim\limits_{r\to 0}\frac {\int_R f(ry)h(y)\,dy}{\int_R g(ry)h(y)dy}=L$ Let $f$ and $g$ be functions defined in a neighborhood of $0$ in $\Bbb R$, such that   $g(x)\neq 0$ in this neighborhood. 
Prove that for all $L\in[-\infty,+\infty]$ and for all non-negative functions $h$ with compact support such that $\int_{\Bbb R} h(y) \,\mathrm dy=1$:
$$\lim_{x\to 0}\frac{f(x)}{g(x)}=L\quad\text{implies that}\quad \lim_{r\to 0}\frac {\int_{\Bbb R} f(ry) h(y)\,\mathrm dy}{\int_{\Bbb R} g(ry)h(y)\,\mathrm dy}=L$$
 A: Let $[-K,K]$ denote some bounded interval containing the support of $h$. Assume first  that $L$ is finite, then for every $u\lt L\lt v$, there exists some $z$ such that $u\lt f(t)/g(t)\lt v$ for every $|t|\lt z$. Add to the hypotheses in the post the hypothesis (H):

(H) $g(t)\gt0$ for every $|t|$ small enough. 

Then, for every $|t|\lt z$, $ug(t)\lt f(t)\lt vg(t)$. Thus, for every $|r|\lt z/K$ and for every real number $t$,
$ug(rt)h(t)\leqslant f(rt)h(t)\leqslant vg(rt)h(t)$ (this uses the hypothesis that $h\geqslant0$ everywhere). Integrating this and using once again the nonnegativity of $g$ and $h$, one sees that the ratio
$$
A(r)=\frac{\int_\mathbb Rf(rt)h(t)\mathrm dt}{\int_\mathbb Rg(rt)h(t)\mathrm dt},
$$
is such that $u\leqslant A(r)\leqslant v$ for every $|r|\leqslant z/K$. Since $u$ and $v$ can be as close to $L$ as one wants, this proves the claim that $A(r)\to L$ when $r\to0$. If $L=+\infty$ or $L=-\infty$, copy this proof, replacing $(u,v)$ by $w$ large enough and of the sign of $L$.
For a counterexample when (H) fails, consider $h=\mathbf 1_{[-1/2,1/2]}$ and $f(t)=t+t^2$ and $g(t)=t+t^4$ if $t\ne0$. Then $L=1$ and
$$
\int_\mathbb Rf(rt)h(t)\mathrm dt=\frac{r^2}{12},\qquad
\int_\mathbb Rg(rt)h(t)\mathrm dt=\frac{r^4}{80},
$$
hence $A(r)\to+\infty$ when $r\to0$. To get the limit $A(r)\to0$ when $r\to0$, exchange the functions $f$ and $g$.
