Where am I wrong in the following limit? We have this function: $f(x)=\frac{2x+3}{x+2}$ and we need to find this: $$\lim _{x\to \infty \:}\frac{\int _x^{2x}f(t)\,dt}{x}$$
Now I will tell how I solved this: I suppose that $$\int _x^{2x} f(t) \, dt=\left.\vphantom{\frac11} F(x)\right|^{2x}_{x}=F(2x)-F(x)$$ and after apply L'Hospital in our limit and we obtain: $$\lim _{x\to \infty } (f(2x)-f(x))=0$$ The problem is that in my book they say this limit is equal with $2$. Where am I wrong?
 A: For any $c \in (x,2x)$ we have 
$$\int_{x}^{2x} f(t) dt= \int_{x}^{c} f(t) dt +\int_{c}^{2x} f(t) dt= \int_{c}^{2x} f(t) - \int_{c}^{x} f(t) dt $$
Then using L'hospital, the Fundamental Theorem of Calculus and the Chain Rule we have 
$$\lim_{x\to \infty} \frac{\color{red}2f(2x) - f(x)}{1} = \lim_{x\to \infty} 2\bigg(\frac{4x + 3}{2x  +2}\Bigg) - \frac{2x + 3}{x+2}$$
And you take it from here.
A: $\displaystyle \int_{x}^{2x} f(t)dt = \left(2t-\ln(t+2)\right)|_{x}^{2x}=2x-\ln(2x+2)+\ln(x+2)\to \displaystyle \lim_{x\to \infty} \dfrac{\displaystyle \int_{x}^{2x} f(t)dt}{x} = \displaystyle \lim_{x\to \infty} \dfrac{2x-\ln(2x+2)+\ln(x+2)}{x}= 2 - \displaystyle \lim_{x\to \infty} \dfrac{\ln(2x+2)}{x}+ \displaystyle \lim_{x\to \infty} \dfrac{\ln(x+2)}{x} = 2 - 0 + 0 = 2$
A: Mean value theorem in $[x,2x]$.  
There is a $z\in (x,2x)$ with $F'(z)=\frac{F(2x)-F(x)}{2x-x}$ which means
$$\frac{2z+3}{z+2}=\frac{\int _x^{2x}f\left(t\right)dt\:}{x}$$
But as $x \to +\infty \Rightarrow z\to+\infty$  the limit of $\frac{2z+3}{z+2}$ equals $2$. 
This is another way to find a solution in many cases where the limit involves this type of functions.
A: Generalization (this is what's really going on): If $f$ is integrable on every bounded subinterval of some $[a,\infty),$ and $\lim_{x\to \infty} f(x)=L,$ then $\lim_{x\to \infty}\frac{1}{x}\int_x^{2x}f(t)\,dt = L.$
