<p>This can be achieved with <a href="https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule" rel="nofollow noreferrer">L'Hopital's rule</a>, which says
<span class="math-container">$$\lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow0}\frac{f'(x)}{g'(x)}.$$</span></p>

<hr>

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

<p>Let
<span class="math-container">$$f(\epsilon)=\int_{l-\epsilon}^{l+\epsilon}h(x,\epsilon)dx \iff f'(\epsilon)=h(l-\epsilon,\epsilon)+h(l+\epsilon,\epsilon)+\int_{l-\epsilon}^{l+\epsilon}\frac{\partial h(x,\epsilon)}{\partial \epsilon}dx$$</span></p>

<p>and
<span class="math-container">$$g(\epsilon)=2\epsilon\iff g'(\epsilon)=2.$$</span></p>

<p>Then, using L'Hopital's rule:
<span class="math-container">$$\lim_{\epsilon\rightarrow0}\frac{f(\epsilon)}{g(\epsilon)}=\lim_{\epsilon\rightarrow0}\frac{f'(\epsilon)}{g'(\epsilon)}=\frac{h(l,0)+h(l,0)}{2}=h(l,0).$$</span></p>

<hr>

<p>A final word on 'intuition'. You reason, correctly, that the integral goes to zero with <span class="math-container">$\epsilon$</span>. But the problem is that the denominator, <span class="math-container">$2\epsilon$</span> also goes to zero. So, to know what the overall expression does as <span class="math-container">$\epsilon$</span> approaches zero, we need to know whether the numerator or the denominator approaches zero faster. In other words, for <span class="math-container">$\epsilon\rightarrow0$</span> 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.</p>

CC BY-SA 4.0
2018-10-15T20:05:28.640Z
That was great but my calculus is rusty. In your original expression for the integral, shouldn't the first term have a minus in front of it ? and where does the third term come from ? I thought that when you take the derivative of the integral, you just evaluate it at both limits. I'm not familiar with the derivative term. Or maybe I used to be and forgot. Thanks.
mark leeds
CC BY-SA 4.0
2018-10-15T20:18:51.957Z
@markleeds: The first term has no minus sign: when the lower limit gets smaller (because $\epsilon$ increases) the value of the integral increases. In essence, there would be a minus sign that comes from the fact that an increase in the lower limit causes the value of the integral to shrink, and a second minus sign that comes from the fact that the lower limit is decreasing in $\epsilon$. The two minus signs cancel. This is basically just the chain rule: $$\frac{\partial}{\partial x}\int_{f(x)}^{c} g(z) dz=f'(x)\frac{\partial}{\partial f(x)}\int_{f(x)}^{c} g(z) dz=-f'(x)g(f(x)).$$
Ubiquitous
CC BY-SA 4.0
2018-10-15T20:27:25.307Z
@markleeds  When we differentiate, we are asking how does a (small) change in $\epsilon$ change the value of the integral. It does so through two effects. Firstly, the limits of the integral change, which is the first two terms. Secondly, the integrand itself changes (that's the $\partial h/\partial \epsilon$ bit), and we have to apply this change along the entire length of the interval over which we are integrating, which is why the $\partial h/\partial \epsilon$ appears inside an integral.
Ubiquitous
CC BY-SA 4.0
2018-10-16T20:59:04.747Z
It's really appreciated. I will print out and go over carefully. Thanks for taking the time for the detailed explanation. I've noticed that people on this list are quite generous and helpful.
mark leeds
CC BY-SA 4.0
2018-10-16T21:00:47.313Z
@markleeds if you need any more help with integration, you can get specialist answers over at math stack exchange.
Ubiquitous
CC BY-SA 4.0
2018-10-16T21:05:36.360Z
probably not in this lifetime aside from trying to follow the integration that's done on this list !!!!!! All the best.
mark leeds
CC BY-SA 4.0
2018-10-15T12:21:42.737Z
2018-10-15T15:52:46.897Z
2018-10-15T15:52:46.897Z
Ubiquitous
<p>I have one question in the proof for section 4.1. in Cole, Mailath, and Postlewaite (2001). 
<span class="math-container">$$\lim_{\varepsilon \to 0}\frac{1}{2\varepsilon}\int_{\overline{l}-\varepsilon}^{\overline{l}+\varepsilon} v(\beta(i) + \delta, \sigma(i+\beta^{-1}(\beta(i) +\delta) - \overline{l}-\varepsilon))-v(\beta(i), s(i))di$$</span></p>

<p>They say the above limit is equal to :
<span class="math-container">$$v(\beta(\overline{l}+\delta, \sigma(\tilde{l}))-v(\beta(\overline{l}, s(\overline{l})),$$</span> for <span class="math-container">$\tilde{l} = \beta^{-1}(\beta(\overline{l})+\delta).$</span></p>

<p>I am not sure how do they this. I think if <span class="math-container">$\varepsilon \to 0$</span>, the term inside the integral goes to 0. But, the answer they give us is to substitute <span class="math-container">$\overline{l}$</span> for <span class="math-container">$i$</span> in the integral. I appreciate if you give some help to figure this out. </p>

<microeconomics><theory>
need help from theorists: proof in Cole, Mailath, and Postlewaite (2001)
shk910
# need help from theorists: proof in Cole, Mailath, and Postlewaite (2001)

(Asked by: username_0 on 2018-10-15)

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.

**Tags:** microeconomics, theory

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## 1 answers

### Answer by username_1 on 2018-10-15. Score: 3

This can be achieved with [L'Hopital's rule](https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_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.

> **Comments:**
> **username_2**: That was great but my calculus is rusty. In your original expression for the integral, shouldn't the first term have a minus in front of it ? and where does the third term come from ? I thought that when you take the derivative of the integral, you just evaluate it at both limits. I'm not familiar with the derivative term. Or maybe I used to be and forgot. Thanks.  
> **username_1**: @markleeds: The first term has no minus sign: when the lower limit gets smaller (because $\epsilon$ increases) the value of the integral increases. In essence, there would be a minus sign that comes from the fact that an increase in the lower limit causes the value of the integral to shrink, and a second minus sign that comes from the fact that the lower limit is decreasing in $\epsilon$. The two minus signs cancel. This is basically just the chain rule: $$\frac{\partial}{\partial x}\int_{f(x)}^{c} g(z) dz=f'(x)\frac{\partial}{\partial f(x)}\int_{f(x)}^{c} g(z) dz=-f'(x)g(f(x)).$$  
> **username_1**: @markleeds  When we differentiate, we are asking how does a (small) change in $\epsilon$ change the value of the integral. It does so through two effects. Firstly, the limits of the integral change, which is the first two terms. Secondly, the integrand itself changes (that's the $\partial h/\partial \epsilon$ bit), and we have to apply this change along the entire length of the interval over which we are integrating, which is why the $\partial h/\partial \epsilon$ appears inside an integral.  
> **username_2**: It's really appreciated. I will print out and go over carefully. Thanks for taking the time for the detailed explanation. I've noticed that people on this list are quite generous and helpful.  
> **username_1**: @markleeds if you need any more help with integration, you can get specialist answers over at math stack exchange.  
> **username_2**: probably not in this lifetime aside from trying to follow the integration that's done on this list !!!!!! All the best.  

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