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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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    from question isn't it clear that $G(a)=\frac{1}{a}\int_{0}^{a}f(x)dx$?2012-06-01

2 Answers 2

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$\frac1a\int_0^af(x)\,\mathrm{d}x$ 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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    @DonAntonio +1 for your answer2012-06-05
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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

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    I'm not sure it matters that much, without any more info, *how* was defined the derivative of the exponential: **if** we can assume the usual stuff about derivatives and if we can use L'H then all is fair play, imo, though I can see your point from an educative-logical stand. I'd love to continue this convo but I don't know where...meta, perhaps? And how?2012-06-01