need help from theorists: proof in Cole, Mailath, and Postlewaite (2001) I have one question in the proof for section 4.1. in Cole, Mailath, and Postlewaite (2001). 
$$\lim_{\varepsilon \to 0}\frac{1}{2\varepsilon}\int_{\overline{l}-\varepsilon}^{\overline{l}+\varepsilon} v(\beta(i) + \delta, \sigma(i+\beta^{-1}(\beta(i) +\delta) - \overline{l}-\varepsilon))-v(\beta(i), s(i))di$$

They say the above limit is equal to :
$$v(\beta(\overline{l}+\delta, \sigma(\tilde{l}))-v(\beta(\overline{l}, s(\overline{l})),$$ for $\tilde{l} = \beta^{-1}(\beta(\overline{l})+\delta).$

I am not sure how do they this. I think if $\varepsilon \to 0$, the term inside the integral goes to 0. But, the answer they give us is to substitute $\overline{l}$ for $i$ in the integral. I appreciate if you give some help to figure this out.
This can be achieved with L'Hopital's rule, which says
$$\lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow0}\frac{f'(x)}{g'(x)}.$$



The result from Cole, Mailath, and Postlewaite holds fairly generally and is not special to the particular functions in their integrand.

Let
$$f(\epsilon)=\int_{l-\epsilon}^{l+\epsilon}h(x,\epsilon)dx \iff f'(\epsilon)=h(l-\epsilon,\epsilon)+h(l+\epsilon,\epsilon)+\int_{l-\epsilon}^{l+\epsilon}\frac{\partial h(x,\epsilon)}{\partial \epsilon}dx$$

and
$$g(\epsilon)=2\epsilon\iff g'(\epsilon)=2.$$

Then, using L'Hopital's rule:
$$\lim_{\epsilon\rightarrow0}\frac{f(\epsilon)}{g(\epsilon)}=\lim_{\epsilon\rightarrow0}\frac{f'(\epsilon)}{g'(\epsilon)}=\frac{h(l,0)+h(l,0)}{2}=h(l,0).$$



A final word on 'intuition'. You reason, correctly, that the integral goes to zero with $\epsilon$. But the problem is that the denominator, $2\epsilon$ also goes to zero. So, to know what the overall expression does as $\epsilon$ approaches zero, we need to know whether the numerator or the denominator approaches zero faster. In other words, for $\epsilon\rightarrow0$ are we dividing a small number by a very very small number, or are we dividing a very very small number by a small number? This, roughly speaking, is the logic behind L'Hopital's rule.