<p>Consider the function <span class="math-container">$f(K,L) = [\alpha K^\rho + \beta L^\rho]^{1/\rho}$</span>. We want to evaluate the limit of <span class="math-container">$f$</span> when <span class="math-container">$\rho \to 0$</span>.</p>
<p><span class="math-container">$$
\lim_{\rho \to 0} f(K,L) = \lim_{\rho \to 0} [\alpha K^\rho + \beta L^\rho]^{1/\rho}
$$</span>
It will turn out to be easier to evaluate <span class="math-container">$\lim_{\rho \to 0} \ln(f(K,L))$</span>:
<span class="math-container">$$
\lim_{\rho \to 0} \ln(f(K,L)) = \lim_{\rho \to 0} \frac{\ln(\alpha K^\rho + \beta L^\rho)}{\rho}
$$</span>
As this evaluates to <span class="math-container">$\frac{0}{0}$</span> we use l'Hôspital's rule:
<span class="math-container">$$
\begin{align*}
\lim_{\rho \to 0} \ln(f(K,L)) &amp;= \lim_{\rho \to 0} \frac{\alpha K^\rho \ln(K) + \beta L^\rho \ln L}{\alpha K^\rho + \beta L^\rho}\\
&amp; = \frac{\alpha \ln(K) + \beta \ln(L)}{\alpha + \beta}\\
&amp; = \frac{\alpha}{\alpha + \beta} \ln(K) + \frac{\beta}{\alpha + \beta}\ln(L).
\end{align*}
$$</span></p>
<p>So taking exponents again gives:
<span class="math-container">$$
\begin{align*}
\lim_{\rho \to 0} f(K,L) &amp;= K^{\frac{\alpha}{\alpha + \beta}} L^{\frac{\beta}{\alpha + \beta}}
\end{align*}
$$</span>
Notice that the factors <span class="math-container">$\frac{\alpha}{\alpha + \beta}$</span> and <span class="math-container">$\frac{\beta}{\alpha + \beta}$</span> sum to one, which implies CRS.</p>

CC BY-SA 4.0
2022-05-16T07:22:11.357Z
2022-05-16T07:22:11.357Z
tdm
<p>Say a CES function is that <span class="math-container">$$Y = A\left[\alpha K^{\rho}+ \beta L^{\rho}\right]^{\frac{1}{\rho}}$$</span>. Clearly this function is constant return to scale whatever the values of <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> take.</p>
<p>Then we know that when <span class="math-container">$\rho=0$</span>, this function goes to the CD case: <span class="math-container">$Y=AK^\alpha L^\beta$</span>. Now it seems only the case where <span class="math-container">$\alpha+\beta=1$</span> satisfies CRS.</p>
<p>What's the intuition of this distinguished result? I am curious because usually in the literature we suppress share parameters and only set say <span class="math-container">$Y = A\left[K^{\rho}+ L^{\rho}\right]^{\frac{1}{\rho}}$</span> or <span class="math-container">$Y=\left[\int_{\Omega}\left(x_{\omega}\right)^{1-\frac{1}{\sigma}} d \omega\right]^{\frac{\sigma}{\sigma-1}}$</span>.</p>
<hr />
<p><strong>Update 2023/8/15</strong></p>
<p>I am reviewing the question and the answer and find it is a little bit more complicated than my original thought.</p>
<p>The answer by @tdm below uses l'Hôspital to get the limitation, while I now notice that the l'Hôspital law itself presumes that <span class="math-container">$\alpha + \beta = 1$</span>. So I don't think this is a prove that when you have arbitrary number of <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span>, and when <span class="math-container">$\rho \rightarrow 0$</span>, you will get a CD production function.</p>
<p>The comment by @walrasian-auctioneer mentions that you can tweak the original production function so that <span class="math-container">$$Y=A\left[\alpha K^\rho+\beta L^\rho\right]^{\frac{1}{\rho}} = A'\left[\alpha' K^\rho+  \beta' L^\rho\right]^{\frac{1}{\rho}} $$</span>, where <span class="math-container">$A' = A(\alpha + \beta)^{\frac{1}{\rho}} $</span>, <span class="math-container">$\alpha' = \frac{\alpha}{\alpha + \beta}$</span>, <span class="math-container">$\beta' = \frac{\beta}{\alpha + \beta}$</span>, and now <span class="math-container">$\alpha' + \beta' = 1$</span>. However note that <span class="math-container">$A'$</span> is not a constant but also a function of <span class="math-container">$\rho$</span>, and thus if we put it into limitation <span class="math-container">$$\lim _{\rho \rightarrow 0} \ln (f(K, L)) = \ln A + \lim _{\rho \rightarrow 0} \left(  \frac{\ln (\alpha+\beta)}{\rho} + \frac{
 \ln \left(\alpha' K^\rho+\beta' L^\rho\right)}{\rho} \right) $$</span>.
The second term in the bracket has been proved as a constant <span class="math-container">$\alpha'\ln (K)+ \beta' \ln (L)$</span>, however the first term goes to positive or negative infinity if <span class="math-container">$\alpha + \beta \neq 1$</span>.</p>
<p>As a result, my current conclusion is that while CES is always constant return to scale (CRS) by its definition without any restriction on the value of <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span>, we must have <span class="math-container">$\alpha + \beta = 1$</span> to transform the CES function into a CD function in the limitation. Please correct me if I am wrong, and any further intuitive explanations are extremely welcomed.</p>

CC BY-SA 4.0
2022-05-14T15:19:56.657Z
Since its CRS, just multiply all inputs by $\frac{1}{(\alpha + \beta)}$ and rescale everything, so when you take it to the Cobb-Douglas limit, the fact that exponents sum to 1 always holds.
Walrasian Auctioneer
<mathematical-economics><production-function><ces-function>
Confusing on the CRS Property of CES Function
Alalalalaki
# Confusing on the CRS Property of CES Function

(Asked by: username_0 on 2022-05-14)

Say a CES function is that $$Y = A\left[\alpha K^{\rho}+ \beta L^{\rho}\right]^{\frac{1}{\rho}}$$. Clearly this function is constant return to scale whatever the values of $\alpha$ and $\beta$ take.

Then we know that when $\rho=0$, this function goes to the CD case: $Y=AK^\alpha L^\beta$. Now it seems only the case where $\alpha+\beta=1$ satisfies CRS.

What's the intuition of this distinguished result? I am curious because usually in the literature we suppress share parameters and only set say $Y = A\left[K^{\rho}+ L^{\rho}\right]^{\frac{1}{\rho}}$ or $Y=\left[\int\_{\Omega}\left(x\_{\omega}\right)^{1-\frac{1}{\sigma}} d \omega\right]^{\frac{\sigma}{\sigma-1}}$.

---

**Update 2023/8/15**

I am reviewing the question and the answer and find it is a little bit more complicated than my original thought.

The answer by @username_2 below uses l'Hôspital to get the limitation, while I now notice that the l'Hôspital law itself presumes that $\alpha + \beta = 1$. So I don't think this is a prove that when you have arbitrary number of $\alpha$ and $\beta$, and when $\rho \rightarrow 0$, you will get a CD production function.

The comment by @username_1 mentions that you can tweak the original production function so that $$Y=A\left[\alpha K^\rho+\beta L^\rho\right]^{\frac{1}{\rho}} = A'\left[\alpha' K^\rho+ \beta' L^\rho\right]^{\frac{1}{\rho}} $$, where $A' = A(\alpha + \beta)^{\frac{1}{\rho}} $, $\alpha' = \frac{\alpha}{\alpha + \beta}$, $\beta' = \frac{\beta}{\alpha + \beta}$, and now $\alpha' + \beta' = 1$. However note that $A'$ is not a constant but also a function of $\rho$, and thus if we put it into limitation $$\lim \_{\rho \rightarrow 0} \ln (f(K, L)) = \ln A + \lim \_{\rho \rightarrow 0} \left( \frac{\ln (\alpha+\beta)}{\rho} + \frac{
\ln \left(\alpha' K^\rho+\beta' L^\rho\right)}{\rho} \right) $$.
The second term in the bracket has been proved as a constant $\alpha'\ln (K)+ \beta' \ln (L)$, however the first term goes to positive or negative infinity if $\alpha + \beta \neq 1$.

As a result, my current conclusion is that while CES is always constant return to scale (CRS) by its definition without any restriction on the value of $\alpha$ and $\beta$, we must have $\alpha + \beta = 1$ to transform the CES function into a CD function in the limitation. Please correct me if I am wrong, and any further intuitive explanations are extremely welcomed.

**Tags:** mathematical-economics, production-function, ces-function

> **Comments:**
> **username_1**: Since its CRS, just multiply all inputs by $\frac{1}{(\alpha + \beta)}$ and rescale everything, so when you take it to the Cobb-Douglas limit, the fact that exponents sum to 1 always holds.  

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## 1 answers

### Answer by username_2 on 2022-05-16. Score: 8

Consider the function $f(K,L) = [\alpha K^\rho + \beta L^\rho]^{1/\rho}$. We want to evaluate the limit of $f$ when $\rho \to 0$.

$$
\lim\_{\rho \to 0} f(K,L) = \lim\_{\rho \to 0} [\alpha K^\rho + \beta L^\rho]^{1/\rho}
$$
It will turn out to be easier to evaluate $\lim\_{\rho \to 0} \ln(f(K,L))$:
$$
\lim\_{\rho \to 0} \ln(f(K,L)) = \lim\_{\rho \to 0} \frac{\ln(\alpha K^\rho + \beta L^\rho)}{\rho}
$$
As this evaluates to $\frac{0}{0}$ we use l'Hôspital's rule:
$$
\begin{align\*}
\lim\_{\rho \to 0} \ln(f(K,L)) &= \lim\_{\rho \to 0} \frac{\alpha K^\rho \ln(K) + \beta L^\rho \ln L}{\alpha K^\rho + \beta L^\rho}\\
& = \frac{\alpha \ln(K) + \beta \ln(L)}{\alpha + \beta}\\
& = \frac{\alpha}{\alpha + \beta} \ln(K) + \frac{\beta}{\alpha + \beta}\ln(L).
\end{align\*}
$$

So taking exponents again gives:
$$
\begin{align\*}
\lim\_{\rho \to 0} f(K,L) &= K^{\frac{\alpha}{\alpha + \beta}} L^{\frac{\beta}{\alpha + \beta}}
\end{align\*}
$$
Notice that the factors $\frac{\alpha}{\alpha + \beta}$ and $\frac{\beta}{\alpha + \beta}$ sum to one, which implies CRS.


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