Confusing on the CRS Property of CES Function

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.

• 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. Commented May 14, 2022 at 15:19

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.