I would like some assistance with a problem that I have showing a CES function is also a Cobb-Douglass utility function.

Question: we have a CES function: $Y=A[\alpha K^{((1-\sigma)/\sigma))}+(1-\alpha) L^{(\sigma-1)/(\sigma)}]^{\sigma/(\sigma-1)}$ $\sigma \geq 0$ $\alpha\in [0,1]$ show it is equivalent to Cobb- Douglass when sigma equals 1

Attempted solution:

$Y=$ $A[\alpha K^{((1-\sigma)/\sigma))}+(1-\alpha) L^{(\sigma-1)/(\sigma)}]^{\sigma/(\sigma-1)}$

$\ln Y=$ $ \ln[A[\alpha K^{((1-\sigma)/\sigma))}+(1-\alpha) L^{(\sigma-1)/(\sigma)}]^{\sigma/(\sigma-1)}]$

$\ln Y=$ $ \frac{\sigma \ln [A[\alpha K^{((1-\sigma)/\sigma))}+(1-\alpha) L^{(\sigma-1)/(\sigma)}]}{\sigma-1}$

Then I take the limit of both sides as $\sigma$ approaches 1

$\lim_{\sigma \rightarrow 1} \ln Y= \\ \lim_{\sigma \rightarrow 1}\frac{\sigma \ln [A[\alpha K^{((1-\sigma)/\sigma))}+(1-\alpha) L^{(\sigma-1)/(\sigma)}]}{\sigma-1}$

I am stuck here. I keep getting $\frac{\ln[1]}{0}=\frac{0}{0}$.

My guess is tht I should use L’hops rule, but I am not sure if my function is differentiable. Even then I doubt I could solve such a complex derivative. Any thoughts, advice, etc.? Also, if I can take the derivative how do I do so? (It’s been a while, LOL)

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    $\begingroup$ This seems to be exactly the same problem already solved here just with different notation. Can you please explain why that answer did not solved your problem? Otherwise this would be a duplicate. $\endgroup$ – 1muflon1 Nov 27 '20 at 17:06
  • $\begingroup$ My apologies. I did not see that problem. Unfortunately it still did not answer my question in full. I was unsure if my calculus was correct up to that point. $\endgroup$ – Tony456 Nov 28 '20 at 17:33
  • $\begingroup$ then please explain in your answer why. As far as I can see the two functions are identical once you substitute $\sigma=1/(1+\rho)$ and set k=1. If you think your problem is different or that solution there does not apply here please edit your question and explain why this should be different problem $\endgroup$ – 1muflon1 Nov 28 '20 at 17:35
  • $\begingroup$ That would be why I assumed my calculus was different. I made an error with $\sigma$. Thank you for catching that. $\endgroup$ – Tony456 Nov 28 '20 at 17:37