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 @tdm 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 @walrasian-auctioneer 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.

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    $\begingroup$ 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. $\endgroup$ May 14, 2022 at 15:19

1 Answer 1


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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