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?

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 $0$. 
