$\lim_t \int (F(x t) - F(0)) / xt dx$ and dominated convergence I'm trying to apply the dominated convergence theorem to compute the following limit:
$$\lim_{t\rightarrow 0} \int_0^1 \frac{F(x t) - F(0)}{xt} dx$$
I know that $F$ is differentiable at $0$ and bounded between $0$ and $1$, but I'm struggling to come up with a dominating function for the integrand.
Does differentiability at $0$ somehow help me find a dominating function?
 A: First of all, for all $x\in[0,1]$ we have
$$ \lim\limits_{t\rightarrow 0}\frac{F(xt)-F(0)}{xt}=F'(0) $$
Suppose without loss of generality that $t\leqslant 1$ and let $\delta>0$ such that $\left|\frac{F(u)-F(0)}{u}-F'(0)\right|\leqslant 1$ for all $u\leqslant\delta$ (such a $\delta$ exists because $F$ is differentiable at $0$). Then $\left|\frac{F(xt)-F(0)}{xt}-F'(0)\right|\leqslant 1$ when $xt\leqslant\delta$, that is when $x\leqslant\frac{\delta}{t}$. On the other hand, if $x>\frac{\delta}{t}$, then $\left|\frac{F(xt)-F(0)}{xt}\right|\leqslant\frac{2\|F\|_{\infty,[0,1]}}{\delta}$ because $0\leqslant xt\leqslant 1$. Therefore,
$$ \left|\frac{F(xt)-F(0)}{xt}\right|\leqslant |F'(0)|+1+\frac{2\|F\|_{\infty,[0,1]}}{\delta} $$
for all $x\in[0,1]$ and $t\in[0,1]$. You have your dominating function, thus you can apply the dominated convergence which gives you the limit :
$$ \lim\limits_{t\rightarrow 0}\int_0^1 \frac{F(xt)-F(0)}{xt}dx=F'(0) $$
A: If dominated convergence is not necessary then simply use the substitution $y=xt$
$$\lim_{t\to0}\frac{1}{t}\int_0^t\frac{F(y)-F(0)}{y}\:dy = \lim_{t\to 0}\frac{F(t)-F(0)}{t} \equiv F'(0)$$
