Question 7.12.5: Consider the “constant elasticity of substitution”, or CES, ......
Consider the “constant elasticity of substitution”, or CES, function
F(K,L)=A\;(a K^{-\rho}+(1-a)L^{-\rho})^{-1/\rho} (∗)
where A > 0, K > 0, L > 0, a ∈ (0, 1), and ρ ≠ 0. Keeping A, K, L, and a fixed, apply l’Hôpital’s rule to z = ln[ F(K, L)/A] as ρ → 0 in order to show that F(K, L) converges to the Cobb–Douglas function AK^{a}L^{1−a}.
\underset{x\to\infty}{lim}{\frac{\ln x}{x}}={“\infty/\infty”}=\underset{x\to\infty}{lim}{\frac{1/x}{1}}=0 l’Hôpital’s rule
Step-by-Step
We get
z=\ln\left(a K^{-\rho}+(1-a)L^{-\rho}\right)^{-1/\rho}=-\ln\left(a K^{-\rho}+(1-a)L^{-\rho}\right)/\rho\rightarrow “0/0” ~as~\rho\to0Because (\mathrm{d}/\mathrm{d}\rho)K^{-\rho}=-K^{-\rho}\ln K\mathrm{and}\,(\mathrm{d}/\mathrm{d}\rho)L^{-\rho}=-L^{-\rho}\ln L, applying l’Hôpital’s rule gives
\underset{\rho\to0}{lim}z=\underset{\rho\to0}{lim}\left[{\frac{a K^{-\rho}\ln K+(1-a)L^{-\rho}\ln L}{a K^{-\rho}+(1-a)L^{-\rho}}}\right]\div 1\quad\quad=a\ln K+(1-a)\ln L
\quad\quad=\ln K^{a}L^{1-a}
Hence e^{z} → K^{a}L^{1−a}. By definition of z, it follows that F(K, L) → AK^{a}L^{1−a} as ρ → 0.
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