Limit of Integral with variable limits of integration For a continuous Function $f$, prove that:
$$\lim_{x\to0^+}\int_{x}^{2x} \frac{1}{t} f(t) dt = ln(2)f(0)$$  
I have already concluded that since f is continuous, it's therefore integrable.Moreover I assumed there is a function $F$ 
$$F(x)=\int_{x}^{2x} f(s)ds$$ 
in order to simplify the limit expression by partial integration. Unfortunately that gave me no solution and I am stuck.
 A: hint 
the substitition $$t=ux $$  makes it easier.
$$I=\int_1^2 f (ux)\frac {du}{u} $$
use the second Mean formula  and continuity at $x=0$.
$$I=f (c_x)\int_1^2\frac {du}{u} $$
Done.
A: $$\int_x ^{2x} \frac{1}{t} f(t) dt-\int_x^{2x}\frac{1}{t}f(0)dt=\int_x^{2x}\frac{1}{t}\left(f(t)-f(0)\right)dt$$
Since $f$ is continuous at $0$, given $\epsilon \gt 0$, there is $\delta \gt 0$ such that $|x| \lt \delta \implies |f(x)-f(0)| \lt \frac{\epsilon}{\ln 2}$. For $|x| \lt \delta$, since $|t| \lt |x| \lt \delta$, we have $$\left|\int_x^{2x}\frac{1}{t}\left(f(t)-f(0)\right)dt\right| \le \int_x^{2x}\left|\frac{1}{t}\left(f(t)-f(0)\right)\right|dt\lt \frac{\epsilon}{\ln 2}\int_x^{2x}\frac{1}{t}dt=\epsilon$$
A: you can use epsilons:
since f is continuous, for a given $\epsilon $ , you can find $ \eta $ such that for all $x \leq \eta$ , we have $ f(0)-\epsilon <=f(x) <= f(0) + \epsilon $
Then you integrate this (divided by t) between x and 2x for x sufficiently small and you get a bounding between $ln(2)(f(0) -\epsilon)$ and $ln(2)(f(0) -\epsilon)$
