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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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    What does the fundamental theorem of calculus say in this case?2012-06-01
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    What is $G(0)?$2012-06-01
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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

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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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    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?2012-06-01
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    @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.2012-06-01
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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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    It is rather silly to apply L'Hopital to the definition of derivative right? Why don't you just put $$\lim_{a\to 0}\frac{G(a)-G(0)}{a-0}=G'(0)$$2012-06-01
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    what is $G(0)=?$2012-06-01
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    @Makuasi We're both (confusingly) taking $G$ to be $$G(a)=\int_0^a f(x) dx$$2012-06-01
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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
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    Yes, as I'm saying maybe we should've chosen another symbol, maybe $H(a)$:2012-06-01
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    Indeed, G wasn't a wise choice, but as for using L'Hospital: Nobody said it was the definition of derivative (though it turned to be the same), so L'H is fair play unless otherwise stipulated. Like $\,\lim_{h\to 0}\frac{e^h-1}{h}\,$...this is the derivative of the exponential at zero, right? Yes, but if nobody forbides it then one can use L'H (i.e., if we're assuming derivatives and L'H are already known)2012-06-01
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    @DonAntonio It depends on how you've proven $(e^x)'=e^x$. If the proof is $f(x)=f'(x)f'(0)$ - where you need to prove $(e^h-1)/h\to 1$ (one option is the definition of the logarithm)- then the argument is circular. It turned out to be the same because it was in the first place a difference quotient. As sos440 is suggesting, FTC could've been rightfully used, for example.2012-06-01
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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