Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Could any one help me for this one ?

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

share|improve this question
2  
What does the fundamental theorem of calculus say in this case? –  sos440 Jun 1 '12 at 4:18
    
What is $G(0)?$ –  El Angel Exterminador Jun 1 '12 at 4:38
    
from question isn't it clear that $G(a)=\frac{1}{a}\int_{0}^{a}f(x)dx$? –  El Angel Exterminador Jun 1 '12 at 4:42

2 Answers 2

$\frac1a\int_0^af(x)\,\mathrm{d}x$ 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, $$ \lim_{a\to0}\frac1a\int_0^af(x)\,\mathrm{d}x=f(0) $$

share|improve this answer
    
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
3  
@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

share|improve this answer
5  
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 $G(0)=?$ –  El Angel Exterminador 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 $G(a)=\frac{1}{a}\int_{0}^{a}f(x)dx$? –  El Angel Exterminador 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.