Compute $\lim_{x \to 0^+}x\int_{x}^1 \frac{f(t)}{t^2}dt $ 
If $f$ is integrable on $[0,1]$ and $\displaystyle \lim_{x \to 0^+}f(x)$ exists, compute $\displaystyle \lim_{x \to 0^+}x\int_{x}^1 \dfrac{f(t)}{t^2} dt $.

We can't really use L'Hospital's rule, but for $f(t) = 1, \forall t$, we get the limit to be $1$. How do we compute the limit? 
 A: Since $f$ is integrable, for all $\varepsilon>0$, $$\int_\varepsilon^1 \frac{f(t)}{t^2}\mathrm d t$$ exist, and thus $$\lim_{x\to 0} x\int_\varepsilon^1 \frac{f(t)}{t^2}\mathrm d t=0$$
for all $\varepsilon>0$. Now, let denote $$\lim_{t\to 0}f(t)=:\ell\in\mathbb R.$$
Then, for all $\delta>0$, there is an $\varepsilon>0$ such that $$\ell-\delta<f(t)<\ell+\delta$$ whenever $t<\varepsilon$. Then,
$$(\ell-\delta) x\int_x^\varepsilon \frac{1}{t^2}\mathrm d t\leq x\int_x^\varepsilon \frac{f(t)}{t^2}\mathrm d t\leq (\ell+\delta)x\int_x^\varepsilon \frac{1}{t^2},$$
and the result follow. 
A: Make the substitution $ t=ux  $ and the integral becomes
$$   x \int_x ^1 \frac{f(t)}{t^2} {\rm d} t= \int _1^{\frac{1}{x}} \frac{f(xu)}{u} {\rm d u}. $$
Now change the variable limit to $1/x$ so we need to calculate 
$ \lim _{x \rightarrow \infty} \int _1^{x} \frac{f(u/x)}{u^2} {\rm d u}   .$
Denote $   \lim _{  x\rightarrow 0}f(x) = f(0+)$.
We will show that the limit is equal to $f(0+).$
Notice that $ \int _{1}^\infty \frac{f(0+)}{u^2} {\rm d } u = \lim_{x \rightarrow \infty} \int _{1}^x \frac{f(0+)}{u^2} {\rm d } u =f(0+)  $.
So it suffices to prove that 
$$\lim _{x \rightarrow \infty} \int _1^{x} \frac{f(u/x) -f(0+)}{u^2} {\rm d u}=0 $$
Break this into two integrals namely
$$\int _1^{x} \frac{f(u/x) -f(0+)}{u^2} {\rm d u} =\int _1^{x^{3/4}} \frac{f(u/x) -f(0+)}{u^2} {\rm d u}+\int _{x^{3/4}}^{x} \frac{f(u/x) -f(0+)}{u^2} {\rm d u} $$
For the second integral we have two estimates (the lower bound is not needed for our purposes)
$$0\leq \frac{1}{x^{2}}\int _{x^{3/4}}^{x} |f(u/x) -f(0+)| {\rm d} u\leq \int _{x^{3/4}}^{x} \frac{|f(u/x) -f(0+)|}{u^2} {\rm d u} \leq \frac{1}{x^{6/4}}\int _{x^{3/4}}^{x}  |f(u/x) -f(0+)| {\rm d} u   $$
\begin{align*}  \frac{1}{x^{6/4}} \int _{x^{3/4}}^{x} |f(u/x)|+|f(0+)| { \rm d}u &=  \frac{1}{x^{6/4}}  \int _{x^{-1/4}}^{1} |f(u)| x{ \rm d}u+ \frac{1}{x^{6/4}}  \int _{x^{-1/4}}^{1} |f(0+)| x{ \rm d}u\\ &=\frac{1}{x^{2/4}}  \int _{x^{-1/4}}^{1} |f(u)| { \rm d}u+ \frac{1}{x^{2/4}}( 1- x^{-1/4}) |f(0+)|  \\ &\leq \frac{1}{x^{2/4}}  \int _{0}^{1} |f(u)| { \rm d}u+ \frac{1}{x^{2/4}} |f(0+)|  \rightarrow 0 \end{align*}
Now by the squeeze theorem when $x\rightarrow \infty$ we get $$\int _{x^{3/4}}^{x} \frac{f(u/x) -f(0+)}{u^2} {\rm d u}=0.$$
Now we can find $M$ such that when $x>M$ we get that $ |f(u/x)-f(0+)|<\epsilon$ $\forall u \in [x,x^{3/4}]$.
Which shows that $$\lim _{x\rightarrow \infty}\int _1^{x^{3/4}} \frac{f(u/x) -f(0+)}{u^2} {\rm d u}=0$$
So we conclude.
