<p>The regular method of obtaining <a href="https://math.stackexchange.com/questions/397925/lhopitals-rule-in-deriving-cobb-douglas-function-from-ces-production-function">Cobb-Douglas</a> and <a href="https://math.stackexchange.com/questions/1024020/limit-of-ces-function-as-p-goes-to-infty">Leotief</a> is <a href="http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule" rel="noreferrer">L'Hôpital's rule</a>. </p>

<p>Another methods should be used too. 
Setting $ \gamma=1$ will be return $Q=[a K^{-\rho} +(1-a) L^{-\rho}  ]^{-\frac{1}{\rho}}$ and 
$$Q^{-\rho}=[a K^{-\rho} +(1-a) L^{-\rho}  ]$$
By The total derivative via differentials we will have
$$-\rho Q^{-\rho-1}dQ=- a\rho K^{-\rho-1}dK -(1-a)\rho L^{-\rho-1}dL   $$
With some manupulations our main equation will be obtained.</p>

<p>$$ dQ= a {(\frac{Q}{ K})}^{1+\rho}dK +(1-a){(\frac{Q}{ L})}^{1+\rho}dL   $$</p>

<p><strong>Linear Function</strong> : $\lim_{\rho\to -1}{dQ}\Rightarrow  Q=aK+(1-a)L$</p>

<p><strong>Cobb-Douglas Function</strong> :  $$\lim_{\rho\to 0}{dQ}\Rightarrow  \frac{1}{Q}dQ= a {(\frac{1}{ K})} dK +(1-a){(\frac{1}{ L})} dL$$
Taking the Integral from both side would produce </p>

<p>$$ \int\frac{1}{Q}dQ=  a \int {(\frac{1}{ K})} dK +(1-a)\int{(\frac{1}{ L})} dL$$</p>

<p>$$Q=K^a L^{(1-a)}e^{C}=AK^a L^{(1-a)}$$</p>

<p><strong>Leontief Function</strong>: $\lim_{\rho\to \infty}{dQ}\Rightarrow  min(aK,(1-a)L)$</p>

CC BY-SA 4.0
2018-11-16T05:15:49.313Z
Is taking a differential and integrating "equivalent" to taking a limit? In general, can we take differential and integrate to find a limit? Or is this a special application?
PGupta
CC BY-SA 3.0
2014-12-07T23:45:39.503Z
(+1) I like especially how the Cobb-Douglas function is obtained.
Alecos Papadopoulos
CC BY-SA 3.0
2014-12-08T20:41:40.687Z
Thanks @AlecosPapadopoulos. but I don't know why somebodies make dislike this post yet? I think this type of questions may provide brain storm  at least to me.
Huseyin
CC BY-SA 3.0
2014-12-08T20:56:43.227Z
Strictly speaking Huseyin, they are right: you should have included at least part of your answer in your _question_: "here is my way of doing things, is there some other way?"
Alecos Papadopoulos
CC BY-SA 3.0
2014-12-03T11:00:13.007Z
2014-12-03T14:34:23.190Z
2017-04-13T12:19:37.213Z
Huseyin
<p>The proofs I will present are based on techniques relevant to the fact that the CES production function has the form of a <a href="http://en.wikipedia.org/wiki/Generalized_mean" rel="noreferrer"><em>generalized weighted mean</em></a>.<br>
This was used in the original paper where the CES function was introduced, <a href="http://www.jstor.org/stable/1927286" rel="noreferrer">Arrow, K. J., Chenery, H. B., Minhas, B. S., &amp; Solow, R. M. (1961). Capital-labor substitution and economic efficiency. The Review of Economics and Statistics, 225-250.</a><br>
The authors there referred their readers to the book <a href="http://books.google.gr/books/about/Inequalities.html?id=t1RCSP8YKt8C&amp;redir_esc=y" rel="noreferrer">Hardy, G. H., Littlewood, J. E., &amp; Pólya, G. (1952). Inequalities </a>, chapter $2 $.</p>

<p>We consider the general case
$$Q_k=\gamma[a K^{-\rho} +(1-a) L^{-\rho}  ]^{-\frac{k}{\rho}},\;\; k&gt;0$$</p>

<p>$$\Rightarrow \gamma^{-1}Q_k = \frac 1{[a (1/K^{\rho}) +(1-a) (1/L^{\rho})  ]^{\frac{k}{\rho}}}$$</p>

<p><strong>1) Limit when $\rho \rightarrow \infty$</strong><br>
Since we are interested in the limit when $\rho\rightarrow \infty$ we can ignore the interval for which $\rho \leq0$, and treat $\rho$ as strictly positive.</p>

<p>Without loss of generality, assume $K\geq L \Rightarrow (1/K^{\rho})\leq (1/L^{\rho})$. We also have $K, L &gt;0$. Then we verify that the following inequality holds:</p>

<p>$$(1-a)^{k/\rho}(1/L^{k})\leq  \gamma Q_k^{-1} \leq (1/L^{k}) $$</p>

<p>$$\implies (1-a)^{k/\rho}(1/L^{k})\leq  [a (1/K^{\rho}) +(1-a) (1/L^{\rho})  ]^{\frac{k}{\rho}} \leq (1/L^{k}) \tag{1}$$</p>

<p>by raising throughout to the $\rho/k$ power to get</p>

<p>$$(1-a)(1/L^{\rho}) \leq a (1/K^{\rho}) +(1-a) (1/L^{\rho})  \leq (1/L^{\rho}) \tag {2}$$
which indeed holds, obviously, given the assumptions. Then go back to the first element of $(1)$ and</p>

<p>$$\lim_{\rho\rightarrow \infty} (1-a)^{k/\rho}(1/L^{k}) =(1/L^{k})$$</p>

<p>which sandwiches the middle term in $(1)$ to $(1/L^{k})$ , so </p>

<p>$$\lim_{\rho\rightarrow \infty}Q_k = \frac {\gamma }{1/L^k} = \gamma L^k = {\gamma }\big[\min\{K,L\}\big]^{k} \tag{3}$$</p>

<p>So for $k=1$ <strong>we obtain the basic Leontief production function.</strong>  </p>

<p><strong>2) Limit when $\rho \rightarrow 0$</strong><br>
Write the function using exponential as</p>

<p>$$\gamma^{-1}Q_k=\exp\left\{-\frac k{\rho}\cdot \ln\big[a (K^{\rho})^{-1} +(1-a) (L^{\rho})^{-1}\big]\right\} \tag {4}$$</p>

<p>Consider the first-order Maclaurin expansion (Taylor expansion centered at zero) of the term inside the logarithm, with respect to $\rho$:</p>

<p>$$a (K^{\rho})^{-1} +(1-a) (L^{\rho})^{-1} \\= a (K^{0})^{-1} +(1-a) (L^{0})^{-1} -a (K^{0})^{-2}K^{0}\rho\ln K- (1-a) (L^{0})^{-2}L^{0}\rho\ln L + O(\rho^2) \\$$</p>

<p>$$=1 - \rho a\ln K - \rho(1-a)\ln L+ O(\rho^2) = 1 +\rho \big[\ln K^{-a}L^{-(1-a)}\big]+ O(\rho^2)$$ </p>

<p>Insert this back into $(4)$ and get rid of the outer exponential,</p>

<p>$$\gamma^{-1}Q_k = \left(1 +\rho \big[\ln K^{-a}L^{-(1-a)}\big]+ O(\rho^{2})\right)^{-k/\rho}$$</p>

<p>In case it is opaque, define $r\equiv 1/\rho$ and re-write</p>

<p>$$\gamma^{-1}Q_k = \left(1 +\frac{\big[\ln K^{-a}L^{-(1-a)}\big]}{r}+ O(r^{-2})\right)^{-kr}$$</p>

<p>Now it does look like an expression whose limit at infinity will give us something exponential:</p>

<p>$$\lim_{\rho\rightarrow 0}\gamma^{-1}Q_k = \lim_{r\rightarrow \infty}\gamma^{-1}Q_k = \left(\exp\left\{ \ln K^{-a}L^{-(1-a)}\right\} \right)^{-k}$$</p>

<p>$$\Rightarrow \lim_{\rho\rightarrow 0}Q_k =\gamma\left(K^{a}L^{1-a}\right)^k$$</p>

<p>The degree of homogeneity $k$ of the function is preserved, and if $k=1$ <strong>we obtain the Cobb-Douglas function.</strong>  </p>

<p>It was this last result that made Arrow and Co to call $a$ the "distribution" parameter of the CES function.</p>

CC BY-SA 4.0
2024-09-11T03:50:20.933Z
Why do you use the exponencial function in equation 4? Let $F(\rho)= a (K^\rho)^{-1} + (1-a) (L^\rho)^{-1}$. You can insert the Taylor expansion
$$F(\rho )= 1 +\rho \big[\ln K^{-a}L^{-(1-a)}\big]+ O(\rho^2)$$
direclty into $Q_k=\gamma[a K^{-\rho} +(1-a) L^{-\rho}  ]^{-\frac{k}{\rho}},\;\; k>0$. Moreover, I think the following is wrong $F(\rho) = a (K^{0})^{-1} +(1-a) (L^{0})^{-1} -a (K^{0})^{-2}K^{0}\rho\ln K- (1-a) (L^{0})^{-2}L^{0}\rho\ln L + O(\rho^2)$. More specifically, the exponent squared in $(K^{0})^{-2}$ and $(L^{0})^{-2}$ should not appear.
MAOC
CC BY-SA 4.0
2024-09-12T15:42:58.503Z
@MAOC If I remember correctly, I have followed Hardy et al in the steps. As for the McLaurin expansion, it is with respect to $\rho$, and we differentiate the function, indicatively, $(K^{\rho})^{-1}$. Its derivative is$ (-1)\cdot(K^{\rho})^{-2}\cdot (\partial K^{\rho}/\partial \rho)$, etc
Alecos Papadopoulos
CC BY-SA 4.0
2024-09-12T15:49:59.160Z
@MAOC ...although I agree that it is too pedantic, since $(K^{\rho})^{-1} = K^{-\rho}$. But it is not wrong.
Alecos Papadopoulos
CC BY-SA 4.0
2024-09-12T19:01:36.293Z
Yes, it's not wrong. I had derived it directly and I found this exponent 2 strange. But it's right. Now, honestly, I don't see any need to use the exponential function in equation 4. I found the approach very beautiful and I believe it looks better if we don't use the exponential trick. Maybe I'm not seeing the need for the trick.
MAOC
CC BY-SA 4.0
2024-09-13T17:53:39.163Z
@MAOC As I said I followed a specific approach I foud in a book. It is not necessarily the only approach that leads to the result, since it appears you have found another and apparently more straightforward one.
Alecos Papadopoulos
CC BY-SA 3.0
2014-12-01T22:32:19.527Z
2016-06-07T19:00:50.267Z
2016-06-07T19:00:50.267Z
Alecos Papadopoulos
<p>In most Microeconomics textbooks it is mentioned that the Constant Elasticity of Substitution (CES) production function,
$$Q=\gamma[a K^{-\rho} +(1-a) L^{-\rho}  ]^{-\frac{1}{\rho}}$$</p>

<p>(where the elasticity of substitution is $\sigma = \frac 1{1+\rho},\rho &gt; -1$), has as its limits both the Leontief production function and the Cobb-Douglas one. Specifically,</p>

<p>$$\lim_{\rho\to \infty}Q= \gamma \min \left \{K , L\right\}$$</p>

<p>and</p>

<p>$$\lim_{\rho\to 0}Q= \gamma K^aL^{1-a}$$</p>

<p>But they never provide the mathematical proof for these results.  </p>

<p>Can somebody please provide these proofs?  </p>

<p>Moreover, the above CES function incorporates constant-returns-to-scale (homogeneity of degree one), due to the outside exponent being $-1/\rho$. If it was, say $-k/\rho$, then the degree of homogeneity would be $k$.  </p>

<p>How are the limiting results affected if $k\neq 1$?</p>

CC BY-SA 3.0
2014-11-28T22:27:09.147Z
This seems to be a homework question with no prior effort of solving it, see: http://meta.economics.stackexchange.com/questions/24/how-should-we-deal-with-homework-questions
FooBar
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2014-11-28T22:32:29.760Z
I can't undrastand that why you say it is homework? http://math.stackexchange.com/questions/1024020/limit-of-ces-function-as-p-goes-to-infty may be a refer to you that this is essential economic question with high quality.
Huseyin
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2014-11-28T23:04:57.480Z
It is certainly an on-topic subject, but a _low-quality question_. Even if it is not homework Huseyin, we expect from you to a) Be careful with your notation (you used $\rho$ and $p$) and b) Contribute some thoughts and ways you have tried to solve the problem. We are here to _help people who help themselves_, and not to offer professional services pro bono.
Alecos Papadopoulos
CC BY-SA 3.0
2014-11-28T23:16:14.417Z
@Alecos_Papadopoulos you are r,ghjt. I edit p and rho . But I think it is not low-quality question. We are in private beta vewrsion so we must provide some related questions with its answers untile it goes to public beta. When it would be public any person who have same question should find the answer.
Huseyin
CC BY-SA 3.0
2014-11-28T23:36:14.677Z
So do you want someone to copy and paste the answer from your link to math.se?
Pburg
CC BY-SA 3.0
2014-11-29T04:31:30.187Z
[math.se] do things differently to pretty much the entire rest of the stackexchange network. Only on math.se can you submit problems for other people to solve without showing effort. Please save that sort of question for math.se, not here.
410 gone
CC BY-SA 3.0
2014-11-29T18:54:48.023Z
I want to know that is there other approach for solving it or not? For example I had tried differinciate appraoch to obtain same resualts. I can not undrastand this offensive method for new persons here.
Huseyin
CC BY-SA 3.0
2014-11-30T04:16:08.957Z
When you say "I need to prove" without any indication of why you need to prove it, people are going to assume this is homework.
Steven Landsburg
CC BY-SA 3.0
2014-11-30T16:08:32.037Z
@StevenLandsburg I wrote it after some people marked my question as homework. I am 40 years old, so having a homework is very sweet dream to me. I am interesting about other solution methods of CES functions. I tried it with differencial equations and obtained Cobb Douglas function. So I am interesting to know other possible methods for do it.
Huseyin
CC BY-SA 3.0
2014-11-30T21:23:40.830Z
@Huseyin If the question re-opens, it would be a good idea to post your solution as an answer to your own question. It appears to be different from the three different ways that the scholars that introduced the CES function, mention for this result.
Alecos Papadopoulos
CC BY-SA 3.0
2014-12-02T20:18:26.547Z
@Huseyin Now that the question has been re-opened and an answer has been provided, won't you post your answer for the Cobb-Douglas limit?
Alecos Papadopoulos
<mathematical-economics><production-function><cobb-douglas><leontief>
How can I obtain Leontief and Cobb-Douglas production function from CES function?
Huseyin
# How can I obtain Leontief and Cobb-Douglas production function from CES function?

(Asked by: username_0 on 2014-11-28)

In most Microeconomics textbooks it is mentioned that the Constant Elasticity of Substitution (CES) production function,
$$Q=\gamma[a K^{-\rho} +(1-a) L^{-\rho} ]^{-\frac{1}{\rho}}$$

(where the elasticity of substitution is $\sigma = \frac 1{1+\rho},\rho > -1$), has as its limits both the Leontief production function and the Cobb-Douglas one. Specifically,

$$\lim\_{\rho\to \infty}Q= \gamma \min \left \{K , L\right\}$$

and

$$\lim\_{\rho\to 0}Q= \gamma K^aL^{1-a}$$

But they never provide the mathematical proof for these results.

Can somebody please provide these proofs?

Moreover, the above CES function incorporates constant-returns-to-scale (homogeneity of degree one), due to the outside exponent being $-1/\rho$. If it was, say $-k/\rho$, then the degree of homogeneity would be $k$.

How are the limiting results affected if $k\neq 1$?

**Tags:** mathematical-economics, production-function, cobb-douglas, leontief
> **Comments:**
> **username_1**: This seems to be a homework question with no prior effort of solving it, see: http://meta.economics.stackexchange.com/questions/24/how-should-we-deal-with-homework-questions  
> **username_0**: I can't undrastand that why you say it is homework? http://math.stackexchange.com/questions/1024020/limit-of-ces-function-as-p-goes-to-infty may be a refer to you that this is essential economic question with high quality.  
> **username_2**: It is certainly an on-topic subject, but a _low-quality question_. Even if it is not homework Huseyin, we expect from you to a) Be careful with your notation (you used $\rho$ and $p$) and b) Contribute some thoughts and ways you have tried to solve the problem. We are here to _help people who help themselves_, and not to offer professional services pro bono.  
> **username_0**: @Alecos_Papadopoulos you are r,ghjt. I edit p and rho . But I think it is not low-quality question. We are in private beta vewrsion so we must provide some related questions with its answers untile it goes to public beta. When it would be public any person who have same question should find the answer.  
> **username_3**: So do you want someone to copy and paste the answer from your link to math.se?  
> **username_4**: [math.se] do things differently to pretty much the entire rest of the stackexchange network. Only on math.se can you submit problems for other people to solve without showing effort. Please save that sort of question for math.se, not here.  
> **username_0**: I want to know that is there other approach for solving it or not? For example I had tried differinciate appraoch to obtain same resualts. I can not undrastand this offensive method for new persons here.  
> **username_5**: When you say "I need to prove" without any indication of why you need to prove it, people are going to assume this is homework.  
> **username_0**: @StevenLandsburg I wrote it after some people marked my question as homework. I am 40 years old, so having a homework is very sweet dream to me. I am interesting about other solution methods of CES functions. I tried it with differencial equations and obtained Cobb Douglas function. So I am interesting to know other possible methods for do it.  
> **username_2**: @Huseyin If the question re-opens, it would be a good idea to post your solution as an answer to your own question. It appears to be different from the three different ways that the scholars that introduced the CES function, mention for this result.  
> **username_2**: @Huseyin Now that the question has been re-opened and an answer has been provided, won't you post your answer for the Cobb-Douglas limit?  

---

## 2 answers

### Answer by username_2 on 2014-12-01. Score: 33 | Accepted answer

The proofs I will present are based on techniques relevant to the fact that the CES production function has the form of a [*generalized weighted mean*](http://en.wikipedia.org/wiki/Generalized_mean).  
This was used in the original paper where the CES function was introduced, [Arrow, K. J., Chenery, H. B., Minhas, B. S., & Solow, R. M. (1961). Capital-labor substitution and economic efficiency. The Review of Economics and Statistics, 225-250.](http://www.jstor.org/stable/1927286)  
The authors there referred their readers to the book [Hardy, G. H., Littlewood, J. E., & Pólya, G. (1952). Inequalities](http://books.google.gr/books/about/Inequalities.html?id=t1RCSP8YKt8C&redir_esc=y) , chapter $2 $.

We consider the general case
$$Q\_k=\gamma[a K^{-\rho} +(1-a) L^{-\rho} ]^{-\frac{k}{\rho}},\;\; k>0$$

$$\Rightarrow \gamma^{-1}Q\_k = \frac 1{[a (1/K^{\rho}) +(1-a) (1/L^{\rho}) ]^{\frac{k}{\rho}}}$$

**1) Limit when $\rho \rightarrow \infty$**  
Since we are interested in the limit when $\rho\rightarrow \infty$ we can ignore the interval for which $\rho \leq0$, and treat $\rho$ as strictly positive.

Without loss of generality, assume $K\geq L \Rightarrow (1/K^{\rho})\leq (1/L^{\rho})$. We also have $K, L >0$. Then we verify that the following inequality holds:

$$(1-a)^{k/\rho}(1/L^{k})\leq \gamma Q\_k^{-1} \leq (1/L^{k}) $$

$$\implies (1-a)^{k/\rho}(1/L^{k})\leq [a (1/K^{\rho}) +(1-a) (1/L^{\rho}) ]^{\frac{k}{\rho}} \leq (1/L^{k}) \tag{1}$$

by raising throughout to the $\rho/k$ power to get

$$(1-a)(1/L^{\rho}) \leq a (1/K^{\rho}) +(1-a) (1/L^{\rho}) \leq (1/L^{\rho}) \tag {2}$$
which indeed holds, obviously, given the assumptions. Then go back to the first element of $(1)$ and

$$\lim\_{\rho\rightarrow \infty} (1-a)^{k/\rho}(1/L^{k}) =(1/L^{k})$$

which sandwiches the middle term in $(1)$ to $(1/L^{k})$ , so

$$\lim\_{\rho\rightarrow \infty}Q\_k = \frac {\gamma }{1/L^k} = \gamma L^k = {\gamma }\big[\min\{K,L\}\big]^{k} \tag{3}$$

So for $k=1$ **we obtain the basic Leontief production function.**

**2) Limit when $\rho \rightarrow 0$**  
Write the function using exponential as

$$\gamma^{-1}Q\_k=\exp\left\{-\frac k{\rho}\cdot \ln\big[a (K^{\rho})^{-1} +(1-a) (L^{\rho})^{-1}\big]\right\} \tag {4}$$

Consider the first-order Maclaurin expansion (Taylor expansion centered at zero) of the term inside the logarithm, with respect to $\rho$:

$$a (K^{\rho})^{-1} +(1-a) (L^{\rho})^{-1} \\= a (K^{0})^{-1} +(1-a) (L^{0})^{-1} -a (K^{0})^{-2}K^{0}\rho\ln K- (1-a) (L^{0})^{-2}L^{0}\rho\ln L + O(\rho^2) \\$$

$$=1 - \rho a\ln K - \rho(1-a)\ln L+ O(\rho^2) = 1 +\rho \big[\ln K^{-a}L^{-(1-a)}\big]+ O(\rho^2)$$

Insert this back into $(4)$ and get rid of the outer exponential,

$$\gamma^{-1}Q\_k = \left(1 +\rho \big[\ln K^{-a}L^{-(1-a)}\big]+ O(\rho^{2})\right)^{-k/\rho}$$

In case it is opaque, define $r\equiv 1/\rho$ and re-write

$$\gamma^{-1}Q\_k = \left(1 +\frac{\big[\ln K^{-a}L^{-(1-a)}\big]}{r}+ O(r^{-2})\right)^{-kr}$$

Now it does look like an expression whose limit at infinity will give us something exponential:

$$\lim\_{\rho\rightarrow 0}\gamma^{-1}Q\_k = \lim\_{r\rightarrow \infty}\gamma^{-1}Q\_k = \left(\exp\left\{ \ln K^{-a}L^{-(1-a)}\right\} \right)^{-k}$$

$$\Rightarrow \lim\_{\rho\rightarrow 0}Q\_k =\gamma\left(K^{a}L^{1-a}\right)^k$$

The degree of homogeneity $k$ of the function is preserved, and if $k=1$ **we obtain the Cobb-Douglas function.**

It was this last result that made Arrow and Co to call $a$ the "distribution" parameter of the CES function.

> **Comments:**
> **username_7**: Why do you use the exponencial function in equation 4? Let $F(\rho)= a (K^\rho)^{-1} + (1-a) (L^\rho)^{-1}$. You can insert the Taylor expansion $$F(\rho )= 1 +\rho \big[\ln K^{-a}L^{-(1-a)}\big]+ O(\rho^2)$$ direclty into $Q_k=\gamma[a K^{-\rho} +(1-a) L^{-\rho}  ]^{-\frac{k}{\rho}},\;\; k>0$. Moreover, I think the following is wrong $F(\rho) = a (K^{0})^{-1} +(1-a) (L^{0})^{-1} -a (K^{0})^{-2}K^{0}\rho\ln K- (1-a) (L^{0})^{-2}L^{0}\rho\ln L + O(\rho^2)$. More specifically, the exponent squared in $(K^{0})^{-2}$ and $(L^{0})^{-2}$ should not appear.  
> **username_2**: @MAOC If I remember correctly, I have followed Hardy et al in the steps. As for the McLaurin expansion, it is with respect to $\rho$, and we differentiate the function, indicatively, $(K^{\rho})^{-1}$. Its derivative is$ (-1)\cdot(K^{\rho})^{-2}\cdot (\partial K^{\rho}/\partial \rho)$, etc  
> **username_2**: @MAOC ...although I agree that it is too pedantic, since $(K^{\rho})^{-1} = K^{-\rho}$. But it is not wrong.  
> **username_7**: Yes, it's not wrong. I had derived it directly and I found this exponent 2 strange. But it's right. Now, honestly, I don't see any need to use the exponential function in equation 4. I found the approach very beautiful and I believe it looks better if we don't use the exponential trick. Maybe I'm not seeing the need for the trick.  
> **username_2**: @MAOC As I said I followed a specific approach I foud in a book. It is not necessarily the only approach that leads to the result, since it appears you have found another and apparently more straightforward one.  

---

### Answer by username_0 on 2014-12-03. Score: 20

The regular method of obtaining [Cobb-Douglas](https://math.stackexchange.com/questions/397925/lhopitals-rule-in-deriving-cobb-douglas-function-from-ces-production-function) and [Leotief](https://math.stackexchange.com/questions/1024020/limit-of-ces-function-as-p-goes-to-infty) is [L'Hôpital's rule](http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule).

Another methods should be used too.
Setting $ \gamma=1$ will be return $Q=[a K^{-\rho} +(1-a) L^{-\rho} ]^{-\frac{1}{\rho}}$ and
$$Q^{-\rho}=[a K^{-\rho} +(1-a) L^{-\rho} ]$$
By The total derivative via differentials we will have
$$-\rho Q^{-\rho-1}dQ=- a\rho K^{-\rho-1}dK -(1-a)\rho L^{-\rho-1}dL $$
With some manupulations our main equation will be obtained.

$$ dQ= a {(\frac{Q}{ K})}^{1+\rho}dK +(1-a){(\frac{Q}{ L})}^{1+\rho}dL $$

**Linear Function** : $\lim\_{\rho\to -1}{dQ}\Rightarrow Q=aK+(1-a)L$

**Cobb-Douglas Function** : $$\lim\_{\rho\to 0}{dQ}\Rightarrow \frac{1}{Q}dQ= a {(\frac{1}{ K})} dK +(1-a){(\frac{1}{ L})} dL$$
Taking the Integral from both side would produce

$$ \int\frac{1}{Q}dQ= a \int {(\frac{1}{ K})} dK +(1-a)\int{(\frac{1}{ L})} dL$$

$$Q=K^a L^{(1-a)}e^{C}=AK^a L^{(1-a)}$$

**Leontief Function**: $\lim\_{\rho\to \infty}{dQ}\Rightarrow min(aK,(1-a)L)$

> **Comments:**
> **username_2**: (+1) I like especially how the Cobb-Douglas function is obtained.  
> **username_0**: Thanks @AlecosPapadopoulos. but I don't know why somebodies make dislike this post yet? I think this type of questions may provide brain storm  at least to me.  
> **username_2**: Strictly speaking Huseyin, they are right: you should have included at least part of your answer in your _question_: "here is my way of doing things, is there some other way?"  
> **username_6**: Is taking a differential and integrating "equivalent" to taking a limit? In general, can we take differential and integrate to find a limit? Or is this a special application?  

---

