Let $f$ be a continuous function on [$0, 1$] with $f(0) =1$. Let $ G(a) = 1/a ∫_0^a f(x)\,dx$ then which of the followings are true? Let $f$ be a continuous function on [$0, 1$] with $f(0) =1$. Let $ G(a) = 1/a ∫_0^af(x)\,dx$ then which of the followings are true?  


*

*$\lim_{(a\to 0)} G(a)=1/2$   

*$\lim_{(a\to0)} G(a)=1$  

*$\lim_{(a\to 0)} G(a)=0$  

*The limit $\lim_{(a\to 0)G(a)}$ does not exist.   


I am completely stuck on it. How should I solve this? 
 A: Note that $G(a)$ is the mean (or average) value of the function on the interval $[0,a]$. Here’s an intuitive argument that should help you see what’s going on. The function $f$ is continuous, and $f(0)=1$, so when $x$ is very close to $0$, $f(x)$ must be close to $1$. Thus, for $a$ close to $0$, $f(x)$ should be close to $1$ for every $x\in[0,a]$, and therefore its mean value should also be close to $1$.
From that it should be easy to pick out the right answer, but it would also be a good exercise for you to try to prove that the answer really is right.
A: Hint: Use l'Hospital's rule for $\lim_{a\to 0}G(a)$. We see that $G(a)\to 1$ when $a$ tends to zero.
A: With the Fundamental theorem of calculus, you can express
\begin{align}
&\int_0^a f(x) \, dx = F(a)-F(0)
\\ \Leftrightarrow &\frac1a\int_0^a f(x) \, dx = \frac{F(a)-F(0)}{a}=G(a)
\end{align}
If you now want to evaluate $\lim_{a\rightarrow 0} G(a)$ you get
\begin{align}
\lim_{a\rightarrow 0} G(a) =\lim_{a\rightarrow 0} \frac{F(a)-F(0)}{a}= ?
\end{align}
Does the above formulation remind you of something?
A: Well, if you let $f(x)=\cos{x}$, then $f(0)=1$ and $\int_0^a f(x) dx=\sin{a}$.  We know that $$\lim_{a\to 0}\frac{\sin(a)}{a}=1$$ as a common identity.  So if this is supposed to hold for all continuous functions on $[0,1]$, then statement $2$ must be true.
This helps you determine the answer quickly; proving it, of course, is a different matter.
