Limit of quotient of integrals I was working with the next problem:

Let $f:[1,\infty)\to\mathbb{R}$ be an increasing function. Consider $F(x)=\displaystyle\int_{1}^{x}\dfrac{f(t)}{t}\, dt$. Calculate the limit $$\lim\limits_{x\to 1^{+}} \dfrac{F(x)}{\ln(x)}=\lim\limits_{x\to 1^{+}} \dfrac{\displaystyle\int_{1}^{x}\dfrac{f(t)}{t}\, dt}{\displaystyle\int_{1}^{x}\dfrac{1}{t}\, dt}$$

Clearly, if we evaluate in the limit, we obtain a quotient of the from $\frac{0}{0}$ but L'Hopital's rule can't be used because we don't know if $F(x)$ is derivable. If we take $f(x)=x$ then the limit is equal to $$\lim\limits_{x\to 1^{+}}\dfrac{\displaystyle\int_{1}^{x}\dfrac{1}{t}\, dt}{\ln(x)}=\lim\limits_{x\to 1^{+}}\dfrac{\ln(x)}{\ln(x)}=1$$If $f(x)=nx$ with $n>0$, then $$\lim\limits_{x\to 1^{+}} \dfrac{\displaystyle\int_{1}^{x}\dfrac{nt}{t}\, dt}{\ln(x)}=\lim\limits_{x\to 1^{+}}\dfrac{n(x-1)}{\ln(x)}=\lim\limits_{x\to 1^{+}}\dfrac{n}{\frac{1}{x}}=\lim\limits_{x\to 1^{+}}nx=n$$The last step is by L'Hopital's rule. But $n=f(1)$. Therefore, we claim that

$$\lim\limits_{x\to 1^{+}}\dfrac{F(x)}{\ln(x)}=f(1)$$

How can I prove it? Any hint? Thanks!
 A: Assume $f$ is continuous. Since $f$ is increasing we have $f(1) \leq f(t) \leq f(1+\epsilon)$ for all $1 \leq t \leq 1+\epsilon$. Then
$$\frac{f(1)}{t} \leq \frac{f(t)}{t} \leq \frac{f(1+\epsilon)}{t},$$
and integrating
$$f(1)\ln x = \int_1^x\frac{f(1)}{t}\,dt \leq F(x) \leq \int_1^x\frac{f(1+\epsilon)}{t}\,dt = f(1+\epsilon)\ln x.$$
Assuming $x > 1$ we have $\ln x > 0$, so we can divide by $\ln x$ to get
$$f(1) \leq \frac{F(x)}{\ln x} \leq f(1+\epsilon),$$
and putting $\epsilon \to 0$ gives your claim.
If $f$ is not continuous, then your claim is false. Take for instance
$$f(t) =
\begin{cases} 
0, & t = 1 \\
1+t, & t > 1.
\end{cases}.$$
Then $f(1) = 0$, but
$$\frac{F(x)}{\ln x} = \frac{\int_1^x \frac{1+t}{t}\,dt}{\ln x} = \frac{\ln x+x-1}{\ln x} = 1 + \frac{x-1}{\ln x},$$
whose limit for $x \to 1^+$ is $2$.
Incidentally, this shows a trick to get all results that you like. For $a \in \mathbb R$ take
$$f_a(t) =
\begin{cases} 
0, & t = 1 \\
a-1+t, & t > 1.
\end{cases}.$$
Then
$$\frac{F(x)}{\ln x} = \frac{\int_1^x \frac{a-1+t}{t}\,dt}{\ln x} = \frac{(a-1)\ln x+x-1}{\ln x} = a-1 + \frac{x-1}{\ln x},$$
and the limit of the latter for $x \to 1^+$ is then $a$.
