# real analysis the lime as x approaches epsilon

Let $f(x)$ be a continuous function. Determine $$\lim_{\epsilon \to 0} \dfrac{1}{\epsilon} \int_{a}^{a+\epsilon} f(x) \, dx$$

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Use the fundamental theorem of Calculus... –  Ayman Hourieh Dec 14 '13 at 23:03
Welcome to MSE, sarah. A couple notes: You should tell us what you've tried when you post a question and you should use Latex to enter math. –  Suugaku Dec 14 '13 at 23:08
–  Mhenni Benghorbal Dec 15 '13 at 3:57

Let $$F(x)=\int_a^xf(t)dt$$ the anti-derivative of $f$ that vanishes at $a$ then $$\lim_{\epsilon\to0}\frac 1\epsilon\int_a^{a+\epsilon}f(t)\,dt=\lim_{\epsilon\to0}\frac{F(a+\epsilon)-F(a)}{\epsilon}=F'(a)=f(a)$$

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$\ddot\smile +1$ –  amWhy Dec 15 '13 at 12:57
Hint: Since $f$ is continuous, when $\epsilon$ is very small the integral is approximately $f(a)\epsilon$.
Geometrically, you can see $\int_{a}^{a+\epsilon}f(x)dx$ as a Riemann sum for $f$, from $a$ to $a+ \epsilon$ (which converges/exists, since $f$ is assumed continuous in the interval of integration) , so that you have a small rectangle, with base-length $a+\epsilon-a= \epsilon$, with height $f(a_k)$, for some $a_k$ in $(a,a+\epsilon)$. By continuity of f at $x=a$ , this height will be $f(a)$ as $\epsilon \rightarrow 0$, so that the integral then equals (as a limit /in the limit) $\epsilon f(a)$ . Then, when this integral is divided by $\epsilon$ , the result is $\frac{1}{\epsilon}f(a)(\epsilon)=f(a)$ as $\epsilon \rightarrow 0$ (actually, for this very last step, you only need $\epsilon \neq 0$).