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Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous. Prove that $\lim_{x \rightarrow 0} \frac{1}{x}\int_0^x f(t) dt = f(0)$.

I'm having a little confusion about proving this. So far, it is clear that $f$ is continuous at 0 and $f$ is Riemann integrable. So with that knowledge, I am trying to use the definition of continuity. So $|\frac{1}{x}\int_0^x f(t) dt - f(0)|=|\frac{1}{x}(f(x)-f(0))-f(0)|$. From here, I'm not sure where to go. Any help is appreciated. Thanks in advance.

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If $f$ is continuos, then $\int_0^x f(t)dt$ is an antiderivate of $f$. Try to use that. – Stefan Nov 27 '12 at 18:15
up vote 5 down vote accepted

$\def\e{\varepsilon}\def\abs#1{\left|#1\right|}$As $f$ is continuous at $0$, for $\e > 0$ there is an $\delta > 0$ such that $\abs{f(x) - f(0)} \le \e$ for $\abs x \le \delta$. For these $x$ we have \begin{align*} \abs{\frac 1x \int_0^x f(t)\, dt - f(0)} &= \abs{\frac 1x \int_0^x \bigl(f(t) - f(0)\bigr)\,dt}\\ &\le \frac 1x \int_0^x \abs{f(t) - f(0)}\, dt\\ &\le \frac 1x \int_0^x \e\,dt\\ &= \e \end{align*} So $\abs{f(0) - \frac 1x \int_0^x f(t)\,dt} \le \e$ for $\abs x \le \delta$, as wished.

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As $\,f\,$ is continuous, we get that

$$\int_0^xf(t)\,dt=F(x)-F(0)\;\;,\;\;\text{with}\,\,\,F'(x)=f(x)\,\,(\text{a primitive of}\,\,f\,)\Longrightarrow$$

$$\lim_{x\to 0}\frac{1}{x}\int_0^xf(t)\,dt=\lim_{x\to 0}\frac{F(x)-F(0)}{x}=F'(0)=f(0)$$

Of course, this is just what Stefan conveyed in his comment.

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+1 That's why we have Theorems, so that we don't need to dive into $(\varepsilon,\delta)$-calculus every time. For reference, I would add that this is based on the fundamental theorem of calculus. – yo' Nov 27 '12 at 19:11
Yes @tohecz, yet I must say I upvoted martini's answer since it is so simple, direct and basic. – DonAntonio Nov 27 '12 at 19:23

We show that as $x$ approaches $0$ from the right, our expression approaches $f(0)$. A similar argument deals with the limit from the left. The two arguments can even be combined, at the cost of intelligibility.

For $x\gt 0$, let $M_x$ be the maximum of $f(t)$ in the interval $0\le t\le x$, and let $m_x$ be the minimum of $f(t)$ in this interval. By basic properties of the Riemann integral, we have $$xm_x \le \int_0^x f(t)\,dt \le xM_x,$$ or equivalently $$m_x\le \frac{1}{x}\int_0^x f(t)\,dt \le M_x.$$ Now let $x\to 0$. As $x$ approaches $0$, both $m_x$ and $M_x$ approach $f(0)$, and therefore by Squeezing so does $\frac{1}{x}\int_0^x f(t)\,dt$.

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