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Let $ f \in C[-1,1]$. Evaluate:

$$\lim_{h \to 0} \frac{ \int^{h}_{-h} f(t) \, dt }{h} $$

My attempt:- Using L' Hospital's rule,

$$\lim_{h \to 0} \frac{\frac{d}{dh} \int^{2h}_{0} f(t) \, dt }{1} = \lim_{h \to 0} f(2h) = f(0)$$

Am I correct?

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How did the limits $-h$ and $h$ turn into $0$ and $2h$? Also, when differentiating that integral you forgot to apply the chain rule. – Keshav Srinivasan Dec 26 '13 at 21:48
up vote 1 down vote accepted

L'Hopital's Rule is valid here, and while there are other ways to do the problem, L'Hopital's Rule works fine. Hint: you have the right idea, but you changed the limits of integration for no reason, and failed to apply the Chain Rule. You need to use the Fundamental Theorem of Calculus and the Chain Rule. You get:

$$\frac{d}{dh} \int_{-h}^h f(t)\,dt= f(h)\frac{d}{dh}(h)-f(-h)\frac{d}{dh}(-h)=f(h)+f(-h).$$

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It seems that using l'Hopital for this is circular (or nearly circular). Why not use more elementary means?

Let $\epsilon > 0$. Since $f$ is continuous at zero, there is $\delta > 0$ so that if $|t-0|<\delta$, then $|f(t)-f(0)|<\epsilon$. Now et $0<h<\delta$. Then every $t \in [-h,h]$ satisfies $|t-0|<\delta$, so $$ (f(0)-\epsilon)\cdot 2h \le \int_{-h}^h f(t)\;dt \le (f(0)+\epsilon)\cdot 2h \\ 2(f(0)-\epsilon) \le \frac{1}{h}\int_{-h}^h f(t)\;dt \le 2(f(0)+\epsilon) $$ So we conclude $$ \lim_{h \to 0^+} \frac{1}{h}\int_{-h}^h f(t)\;dt = 2f(0) . $$

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Another approach would be to use the Mean Value Theorem for Integrals:

Since $\int_{-h}^{h} f(t)\, dt= 2hf(c_{h})$ for some $c_{h}$ in $(-h,h)\;\;\;$ (or $(h,-h)$, if $h<0$), $$\lim_{h \to 0} \frac{ \int^{h}_{-h} f(t) \, dt }{h}=\lim_{h\to 0} \frac{2hf(c_{h})}{h}=\lim_{h\to 0}2f(c_{h})=2f(0).$$

$------------------------------------------\\$ Here is a similar argument which does not use the MVT for Integrals:

We can assume that $h>0$, since the function is an even function of h; so

let $M_h$ and $m_h$ be the maximum and minimum values of f on $[-h,h]$.

Then $2hm_h\le \int_{-h}^{h} f(t)\;dt\le 2hM_h\;\;$ and $\displaystyle\;\;2m_h\le \frac{ \int^{h}_{-h} f(t) \, dt }{h}\le 2M_h $ for $h>0$.

Since $\displaystyle \lim_{h\to 0}2m_h=2f(0)$ and $\displaystyle \lim_{h\to 0}2M_h=2f(0)\;\;$ since f is continuous at 0,

$\displaystyle\;\;\;\lim_{h \to 0} \frac{ \int^{h}_{-h} f(t) \, dt }{h}=2f(0)$ by the Squeeze Theorem.

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I don't like that solution. It assumes the axiom of choice, which is a bit of overkill for a calculus problem. – Keshav Srinivasan Dec 26 '13 at 23:22
Let $c_h$ be the least solution ... the set of all solutions is a closed set. So AC is not required. – GEdgar Dec 27 '13 at 21:53

Assuming $\;f\;$ is continuous and thus has a primitive function $\;F\;$ on $\;[-1,1]\;$ , we'd get applying l'Hospital's rule:

$$\lim_{h\to 0}\frac1h\int\limits_{-h}^hf(t)dt=\lim_{h\to 0}\frac{F(h)-F(-h)}h\stackrel{\text{l'H}}=\lim_{h\to 0}(F'(h)+F'(-h))=$$

$$=\lim_{h\to 0}(f(h)+f(-h))=2f(0)$$

and we do not get what you got in this particular case...

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It's probably not a good idea to show the full solution. – Keshav Srinivasan Dec 26 '13 at 22:08
I think in this particular case it is as the question was "Am I correct?" and my particular example shows the OP wasn't. There's no much room for anything else here, imo. – DonAntonio Dec 26 '13 at 22:15
Its my mistake! :( – wannadeleteacct Dec 26 '13 at 22:22

Hint: In your attempt, you just changed the limits of integration without justification, but tree to see whether you can somehow rewrite your integral into a sum or difference of integrals whose lower limits are constant. That way you can easily apply the fundamental theorem of calculus. And don't forget to use the chain rule if the need arises.

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yes,thats right.. oops! I should have taken $\int^h_0$ and $ \int^0_{-h}$.That was a little blunder. – wannadeleteacct Dec 26 '13 at 22:19
@Manasi Or better yet, you could break the integral into the integral from 0 to h minus the integral from 0 to -h. That way you have a constant as the lower limit in both integrals. – Keshav Srinivasan Dec 26 '13 at 23:25
Well yes, thats how I did it when I looked at the posts pointing my mistake. Thanks again! I should not have asked this question in the first place and blundered the way I did.. – wannadeleteacct Dec 27 '13 at 10:18

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