If $f$ is integrable on $[0,1]$, and $\lim_{x\to 0^+}f(x)$ exists, compute $\lim_{x\to 0^{+}}x\int_x^1 \frac{f(t)}{t^2}dt$. If $f$ is integrable on $[0,1]$, and $\lim_{x\to 0}f(x)$ exists, compute $\lim_{x\to 0^{+}}x\int_x^1 \frac{f(t)}{t^2}dt$.
I'm lost about what the value is for this limit in the first place. How can I make a guess for this kind of limit?
 A: Denote the limit by $\ell = \lim_{x \to 0^+} f(x)$. The main observation is that only the values of $f$ near zero is important. Accordingly we decompose the integral as follows:
$$ x \int_{x}^{1} \frac{f(t)}{t^2} \, dt
= \underbrace{x \int_{x}^{1} \frac{f(t) - \ell}{t^2} \, dt}_{=: A(x)}
+ \underbrace{x \int_{x}^{1} \frac{\ell}{t^2} \, dt}_{=:B(x)}. $$ 
We will check that $A(x) \to 0$ and $B(x) \to \ell$ as $x \to 0^+$. Once these are established, we have

$$ \lim_{x\to0^+} x \int_{x}^{1} \frac{f(t)}{t^2} \, dt= \ell = \lim_{x\to0^+} f(x). $$


Step 1. We show that $A(x) \to 0$. To this end, for any $\epsilon > 0$ we choose $\delta > 0$ such that
$$ 0 < x \leq \delta \quad \Rightarrow \quad |f(x) - \ell| \leq \epsilon. $$
Thus for $0 < x < \delta$ we decompose $A(x)$ into 'near-zero part' and 'away-from-zero part' and estimate them separately:
\begin{align*}
|A(x)|
&\leq x \int_{x}^{\delta} \frac{|f(t) - \ell|}{t^2} \, dt + x \int_{\delta}^{1} \frac{|f(t) - \ell|}{t^2} \, dt \\
&\leq x \int_{x}^{\delta} \frac{\epsilon}{t^2} \, dt + x \int_{\delta}^{1} \frac{|f(t) - \ell|}{t^2} \, dt \\
&\leq \epsilon + x \underbrace{\int_{\delta}^{1} \frac{|f(t) - \ell|}{t^2} \, dt}_{\text{constant}}.
\end{align*}
Taking limsup as $x \to 0^+$, we have
$$ \limsup_{x \to 0^+} |A(x)| \leq \epsilon. $$
(Notice here that the limit of $A(x)$ is not a priori clear. That is why we used limsup instead.) Since this is true for any $\epsilon > 0$ and the LHS is independent of $\epsilon$, we conclude that $A(x) \to 0$ as $x \to 0^+$.
Step 2. By direct calculation, $ B(x) = \ell(1 - x) \to \ell$ as $x \to 0^+$.
Combining Step 1 and 2, we confirm the claim and the proof is done. ///
A: To make a guess, first consider the simplest kind of function for which the limit exists, say $f=1$. Then $\lim_{x\to 0^{+}}x\int_x^1 \frac{f(t)}{t^2}dt=\lim_{x\to 0^{+}}x(1/x-1)=1$. 
So clearly, if $\lim_{x\to 0} f(x)=l$, then $\lim_{x\to 0^{+}}x\int_x^1 \frac{f(t)}{t}dt=l$.
Now let's prove this. Let's use the condition of the $\lim_{x\to 0} f(x)=l$.
Given any $\epsilon \gt 0$, there is some $\delta \gt 0$ such that $|f(t)-l|\lt \epsilon$ for $0\lt t\lt \delta$. 
Then
$|x\int_x^1 \frac{f(t)-l}{t^2}dt|\le x\int_x^{\delta} \frac{|f(t)-l|}{t^2}dt+x|\int_{\delta}^1 \frac{f(t)-l}{t^2}dt|\le x\epsilon \int_x^{\delta}\frac{dt}{t^2}+x|\int_{\delta}^1 \frac{f(t)-l}{t^2}dt|$.
Hence we get 
$|x\int_x^1 \frac{f(t)}{t^2}dt+xl-l|\le \epsilon-\frac{\epsilon x}{\delta}+x|\int_{\delta}^1 \frac{f(t)-l}{t^2}dt|$.
Finally, we have
$|x\int_x^1 \frac{f(t)}{t^2}dt-l|=|x\int_x^1 \frac{f(t)}{t^2}dt+xl-l-xl|\le |x\int_x^1 \frac{f(t)}{t^2}dt+xl-l|+|xl|\le \epsilon-\frac{\epsilon x}{\delta}+x|\int_{\delta}^1 \frac{f(t)-l}{t^2}dt|+|xl|$. 
Now in both sides, let $x\to 0^{+}$, then we have
$\lim_{x\to 0^{+}}|x\int_x^1 \frac{f(t)}{t^2}dt-l|\le \epsilon$, and since $\epsilon\gt 0$ is arbitrary, this limit is $0$. 
