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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?

  1. $\lim_{(a\to 0)} G(a)=1/2$
  2. $\lim_{(a\to0)} G(a)=1$
  3. $\lim_{(a\to 0)} G(a)=0$
  4. The limit $\lim_{(a\to 0)G(a)}$ does not exist.

I am completely stuck on it. How should I solve this?

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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.

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Sorry for asking but should I delete my answer while yours and other's answers are in details? – Babak S. Dec 20 '12 at 9:18
@Babak: I don’t see any real reason to do so: I upvoted it. – Brian M. Scott Dec 20 '12 at 9:26
Thanks Brian for supporting others. You are really our kind father here. :-) – Babak S. Dec 20 '12 at 9:31

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.

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L'Hopital to the rescue once again... $\ddot\smile$ – amWhy Apr 15 '13 at 0:24
@amWhy: Amy, I think, it is better he had change his family name to L'Hospital. Cause it really is needed in emergency cases in limits. :D – Babak S. Apr 15 '13 at 7:51

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?

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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.

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