<p><strong>Yes.</strong></p>

<p>Write</p>

<p>$$U(x,y) = (ax^{-c} + by^{-c})^{-\frac{1}{c}} = \exp\left\{\frac {-1}{c}\ln \left(ax^{-c} + by^{-c}\right)\right\} \tag{1}$$</p>

<p>Now <em>if</em> $a+b=1$ then as $c\rightarrow 0$, expression </p>

<p>$$\frac {\ln \left(ax^{-c} + by^{-c}\right)}{-c} \tag{2}$$ </p>

<p>will be an indeterminate form $0/0$ and so we can apply L'Hopital's rule on it to get</p>

<p>$$\frac {1}{-c}\cdot \frac {-ax^{-c}\ln x + by^{-c}\ln y}{ax^{-c} - by^{-c}} \rightarrow \frac {a}{a+b}\ln x + \frac {b}{a+b}\ln y,\;\;\; c\rightarrow 0$$</p>

<p>where we have assumed $a+b=1$.</p>

<p>So (by the uniform continuity of the exponential function)</p>

<p>$$c\rightarrow 0,\;\;\; U(x,y) = \exp\left\{\frac {-1}{c}\ln \left(ax^{-c} + by^{-c} \right) \right\} \rightarrow \exp\left\{a\ln x + b\ln y \right \} = x^ay^b$$.</p>

<p>But if  $a+b\neq 1$, then as $c\rightarrow 0$, the argument of the logarithm in the numerator in eq. $(2)$ would not have been equal to unity, and so the logarithm would not equal zero (which would give us the indeterminate form, and would allow us to use L' Hopital's rule). Instead, eq. $(2)$ would have gone to infinity. So $a+b=1$ is needed in order to arrive at the Cobb-Douglas function.</p>

CC BY-SA 3.0
2016-02-03T18:19:35.787Z
2016-02-03T18:19:35.787Z
Alecos Papadopoulos
<p>Consider the C.E.S. utility function</p>

<p>$$U(x, y) = (ax^{-c} + by^{-c})^{-\frac{1}{c}} $$</p>

<p>Is it true that we must have $a+b=1$ in order to obtain a Cobb-Douglas utility function as $c\rightarrow 0$?</p>

CC BY-SA 3.0
2016-02-04T08:06:06.327Z
Possible duplicate of [How does the limit of $U(x, y) = (ax^{-c} + by^{-c})^{-\frac{1}{c}}$ as c approaches 0 yield the Cobb-Douglas utlity function?](http://economics.stackexchange.com/questions/10497/how-does-the-limit-of-ux-y-ax-c-by-c-frac1c-as-c-approa)
Oliv
CC BY-SA 3.0
2016-02-04T13:30:14.320Z
@Oliv This thread is certainly a variant of the thread you mentioned, and my original intention was to post an answer there just to make even more visible the necessity of $a+b=1$. The reason I had to create a self-Q&A was that the question you mention had been closed and no new answers were accepted by the system.
Alecos Papadopoulos
CC BY-SA 3.0
2016-02-05T19:09:52.443Z
Possible duplicate of [How can I obtain Leontief and Cobb-Douglas production function from CES function?](http://economics.stackexchange.com/questions/361/how-can-i-obtain-leontief-and-cobb-douglas-production-function-from-ces-function)
FooBar
<microeconomics><mathematical-economics><cobb-douglas>
In the C.E.S. utility function do the parameters need to add up to unity to obtain the Cobb-Douglas utility function?
Alecos Papadopoulos
# In the C.E.S. utility function do the parameters need to add up to unity to obtain the Cobb-Douglas utility function?

(Asked by: username_0 on 2016-02-03)

Consider the C.E.S. utility function

$$U(x, y) = (ax^{-c} + by^{-c})^{-\frac{1}{c}} $$

Is it true that we must have $a+b=1$ in order to obtain a Cobb-Douglas utility function as $c\rightarrow 0$?

**Tags:** microeconomics, mathematical-economics, cobb-douglas

> **Comments:**
> **username_1**: Possible duplicate of [How does the limit of $U(x, y) = (ax^{-c} + by^{-c})^{-\frac{1}{c}}$ as c approaches 0 yield the Cobb-Douglas utlity function?](http://economics.stackexchange.com/questions/10497/how-does-the-limit-of-ux-y-ax-c-by-c-frac1c-as-c-approa)  
> **username_0**: @username_1 This thread is certainly a variant of the thread you mentioned, and my original intention was to post an answer there just to make even more visible the necessity of $a+b=1$. The reason I had to create a self-Q&A was that the question you mention had been closed and no new answers were accepted by the system.  
> **username_2**: Possible duplicate of [How can I obtain Leontief and Cobb-Douglas production function from CES function?](http://economics.stackexchange.com/questions/361/how-can-i-obtain-leontief-and-cobb-douglas-production-function-from-ces-function)  

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## 1 answers

### Answer by username_0 on 2016-02-03. Score: 5

**Yes.**

Write

$$U(x,y) = (ax^{-c} + by^{-c})^{-\frac{1}{c}} = \exp\left\{\frac {-1}{c}\ln \left(ax^{-c} + by^{-c}\right)\right\} \tag{1}$$

Now *if* $a+b=1$ then as $c\rightarrow 0$, expression

$$\frac {\ln \left(ax^{-c} + by^{-c}\right)}{-c} \tag{2}$$

will be an indeterminate form $0/0$ and so we can apply L'Hopital's rule on it to get

$$\frac {1}{-c}\cdot \frac {-ax^{-c}\ln x + by^{-c}\ln y}{ax^{-c} - by^{-c}} \rightarrow \frac {a}{a+b}\ln x + \frac {b}{a+b}\ln y,\;\;\; c\rightarrow 0$$

where we have assumed $a+b=1$.

So (by the uniform continuity of the exponential function)

$$c\rightarrow 0,\;\;\; U(x,y) = \exp\left\{\frac {-1}{c}\ln \left(ax^{-c} + by^{-c} \right) \right\} \rightarrow \exp\left\{a\ln x + b\ln y \right \} = x^ay^b$$.

But if $a+b\neq 1$, then as $c\rightarrow 0$, the argument of the logarithm in the numerator in eq. $(2)$ would not have been equal to unity, and so the logarithm would not equal zero (which would give us the indeterminate form, and would allow us to use L' Hopital's rule). Instead, eq. $(2)$ would have gone to infinity. So $a+b=1$ is needed in order to arrive at the Cobb-Douglas function.


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