A question on limit with definite integral

Could any one help me for this one ?

If $f$ is continuous on $[0,1]$ and $f(0)=1$, then $$\lim\limits_{a\to 0}G(a)=\frac{1}{a}\int_{0}^{a}f(x)dx=?$$

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What does the fundamental theorem of calculus say in this case? –  sos440 Jun 1 '12 at 4:18
What is $G(0)?$ –  Bunuelian Trick Jun 1 '12 at 4:38
from question isn't it clear that $G(a)=\frac{1}{a}\int_{0}^{a}f(x)dx$? –  Bunuelian Trick Jun 1 '12 at 4:42

$\frac1a\int_0^af(x)\,\mathrm{d}x$ is the average of $f$ over $[0,a]$. Since $f$ is continuous, as $a\to0$, $f$ on $[0,a]$ is close to $f(0)$. Thus, a good guess would be that $\frac1a\int_0^af(x)\,\mathrm{d}x=f(0)$. Let's add some rigor.

Since $f$ is continuous at $0$, for any $\epsilon>0$, there is a $\delta>0$ so that for all $|x-0|<\delta$, we have $|f(x)-f(0)|<\epsilon$, and then \begin{align} \left|\lim_{a\to0}\frac1a\int_0^af(x)\,\mathrm{d}x-f(0)\right| &=\lim_{a\to0}\frac1a\left|\int_0^a(f(x)-f(0))\,\mathrm{d}x\right|\\ &\le\lim_{a\to0}\frac1a\int_0^a|f(x)-f(0)|\,\mathrm{d}x\\ &\le\lim_{a\to0}\frac1a\int_0^a\epsilon\,\mathrm{d}x\\ &=\epsilon \end{align} Since $\epsilon$ is arbitrary, $$\lim_{a\to0}\frac1a\int_0^af(x)\,\mathrm{d}x-f(0)=0$$ Therefore, $$\lim_{a\to0}\frac1a\int_0^af(x)\,\mathrm{d}x=f(0)$$

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The above is not a way to find the limit as asked by the OP but just a proof that the limit is $\,f(0)\,$...how can we know beforehand this the limit? –  DonAntonio Jun 1 '12 at 14:41
@DonAntonio: sometimes we guess the answer and then prove that guess. In this case, $\frac1a\int_0^af(x)\,\mathrm{d}x$ is the average of $f$ over $[0,a]$. $f$ is continuous, so as $a\to0$, $f(x)$ is close to $f(0)$ for $x\in[0,a]$. Therefore, $\frac1a\int_0^af(x)\,\mathrm{d}x=f(0)$ is a good guess. –  robjohn Jun 1 '12 at 16:33
@DonAntonio +1 for your answer –  srijan Jun 5 '12 at 10:39
Apply L'Hospital: with $G$ a primitive of $f$ in the unit interval,$$\lim_{a\to 0}\frac{1}{a}\int_0^a\,f(x)\,dx=\lim_{a\to 0}\frac{G(a)-G(0)}{a}=\lim_{a\to 0}G'(a)=\lim_{a\to 0}f(a)=1$$ by continuity
It is rather silly to apply L'Hopital to the definition of derivative right? Why don't you just put $$\lim_{a\to 0}\frac{G(a)-G(0)}{a-0}=G'(0)$$ –  Pedro Tamaroff Jun 1 '12 at 4:20
what is $G(0)=?$ –  Bunuelian Trick Jun 1 '12 at 4:36
@Makuasi We're both (confusingly) taking $G$ to be $$G(a)=\int_0^a f(x) dx$$ –  Pedro Tamaroff Jun 1 '12 at 4:38
from question isn't it clear that $G(a)=\frac{1}{a}\int_{0}^{a}f(x)dx$? –  Bunuelian Trick Jun 1 '12 at 4:40
Yes, as I'm saying maybe we should've chosen another symbol, maybe $H(a)$: –  Pedro Tamaroff Jun 1 '12 at 4:45