<p>It is not true that this function is equivalent to the Cobb-Douglas utility function when $c \sim 0$ for any values of $(a,b)$; you have to assume $a+b=1$ for that, i.e. $b=1-a$.</p>

<p>To see why it is true, fix $(x,y)$ and consider the following Taylor expansion of $U(x,y)$ when $c$ gets close to $0$. We have</p>

<p>\begin{align*}
(ax^{-c}+by^{-c})^{-\frac{1}{c}} &amp; = e^{-\frac{1}{c} \ln{[ax^{-c}+by^{-c}]}} \\
&amp; = e^{-\frac{1}{c} \ln{[ae^{-c\ln(x)}+be^{-c \ln(y)}]}} \\
&amp; = e^{-\frac{1}{c} \ln{[a(1-c \ln(x) + o(c)) + b(1-c \ln(y) + o(c))}]} \\
&amp; = e^{-\frac{1}{c} \ln{[a+b - c (a \ln(x)+b \ln(y)) + o(c))]}} \\
\end{align*}
If $a+b&gt;1$, the term in the exponential converges to $-\infty$ when $c \rightarrow 0$ and therefore $U(x,y) \rightarrow 0$. If $a+b&lt;1$, the term converges to $+\infty$ and therefore $U(x,y) \rightarrow +\infty$.</p>

<p>To obtain the convergence towards the Cobb-Douglas function, we must therefore assume $a+b=1$. In that case we have
\begin{align*}
U(x,y) &amp; = e^{-\frac{1}{c} \ln{[1-c(a \ln(x) + (1-a) \ln(y))+o(c)]}} \\
&amp; = e^{-\frac{1}{c} [-c(a \ln(x) + (1-a) \ln(y))+o(c)]} \\
&amp; = e^{a \ln(x) + (1-a) \ln(y) + \frac{o(c)}{c}} \\
&amp; \rightarrow_{c \rightarrow 0} e^{a \ln(x) + (1-a) \ln(y)} \\
&amp; = x^{a} y^{1-a}
\end{align*}
which is the Cobb-Douglas utility function with parameters $(a,1-a)$.</p>

CC BY-SA 3.0
2016-02-04T06:04:56.967Z
How did you get rid of the error term o(c)/c?
bloopton
CC BY-SA 3.0
2016-02-04T07:15:07.397Z
@bloopton By definition of $o(c)$, the fraction $\frac{o(c)}{c}$ converges to $0$ when $c$ tends to $0$.
Oliv
CC BY-SA 3.0
2016-02-03T09:50:30.330Z
2016-02-03T09:50:30.330Z
Oliv
<p>\begin{equation*}
U(x, y) = (ax^{-c} + by^{-c})^{-\frac{1}{c}} 
\end{equation*}</p>

<p>I ask this mainly because after logging both sides of the Utility equation (the first step to proving the assertion, I assume), I am left with:</p>

<p>\begin{equation*}
\lim_{c \rightarrow 0} \dfrac{-\ln(ax^{-c} + by^{-c})}{c}
\end{equation*}
I know that the bottom will go to 0, and I have a feeling that the top will go to 0 to.  However, all I am left with on the top is essentially $a + b$, and for it to go to 0, $a + b = 1$.</p>

<p>How can $a + b = 1$?  Is this the right direction?  What does $a + b = 1$ mean?  Why does $a + b = 1$?</p>

<p>Edit:  And once proven, what does this whole "limit" thing say about the original function?  What is so special about this particular equation such that its limit as $c \rightarrow 0$ is the Cobb Douglas function?</p>

<p>Edit 2:  Upon further research, I have discovered a suspiciously similar function known as the CES.  $a$ and $b$, however, are instead $a$ and $(1-a)$ !!  Now I'm even more confused.  How am I supposed to derive that complementary relationship from this equation?  This is supposed to be consumer theory!</p>

CC BY-SA 3.0
2016-02-03T09:12:53.630Z
Let us clarify something. If
$$
\lim_{x\to 0} f(x) = 0
$$
then
$$
\lim_{x\to 0} \ln(f(x)) = -\infty
$$
not zero, as you seem to assume.
Giskard
CC BY-SA 3.0
2016-02-03T14:48:41.273Z
@denesp Sorry was mistake, I corrected it.  Was supposed to be 1.
bloopton
CC BY-SA 3.0
2016-02-03T14:49:30.517Z
@AlecosPapadopoulos I found that question and that's what confused me.  I'm studying consumer theory right now, not producer theory.
bloopton
CC BY-SA 3.0
2016-02-03T17:36:32.977Z
The functional form is  the same, and so it is purely mathematics -and they are the same mathematics. Write down the math in the post I linked to using the symbols from your consumer theory set up.
Alecos Papadopoulos
<microeconomics><cobb-douglas>
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?
bloopton
# 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?

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

\begin{equation\*}
U(x, y) = (ax^{-c} + by^{-c})^{-\frac{1}{c}}
\end{equation\*}

I ask this mainly because after logging both sides of the Utility equation (the first step to proving the assertion, I assume), I am left with:

\begin{equation\*}
\lim\_{c \rightarrow 0} \dfrac{-\ln(ax^{-c} + by^{-c})}{c}
\end{equation\*}
I know that the bottom will go to 0, and I have a feeling that the top will go to 0 to. However, all I am left with on the top is essentially $a + b$, and for it to go to 0, $a + b = 1$.

How can $a + b = 1$? Is this the right direction? What does $a + b = 1$ mean? Why does $a + b = 1$?

Edit: And once proven, what does this whole "limit" thing say about the original function? What is so special about this particular equation such that its limit as $c \rightarrow 0$ is the Cobb Douglas function?

Edit 2: Upon further research, I have discovered a suspiciously similar function known as the CES. $a$ and $b$, however, are instead $a$ and $(1-a)$ !! Now I'm even more confused. How am I supposed to derive that complementary relationship from this equation? This is supposed to be consumer theory!

**Tags:** microeconomics, cobb-douglas

> **Comments:**
> **username_1**: Let us clarify something. If $$ \lim_{x\to 0} f(x) = 0 $$ then $$ \lim_{x\to 0} \ln(f(x)) = -\infty $$ not zero, as you seem to assume.  
> **username_0**: @denesp Sorry was mistake, I corrected it.  Was supposed to be 1.  
> **username_0**: @username_2 I found that question and that's what confused me.  I'm studying consumer theory right now, not producer theory.  
> **username_2**: The functional form is  the same, and so it is purely mathematics -and they are the same mathematics. Write down the math in the post I linked to using the symbols from your consumer theory set up.  

---

## 1 answers

### Answer by username_3 on 2016-02-03. Score: 10 | Accepted answer

It is not true that this function is equivalent to the Cobb-Douglas utility function when $c \sim 0$ for any values of $(a,b)$; you have to assume $a+b=1$ for that, i.e. $b=1-a$.

To see why it is true, fix $(x,y)$ and consider the following Taylor expansion of $U(x,y)$ when $c$ gets close to $0$. We have

\begin{align\*}
(ax^{-c}+by^{-c})^{-\frac{1}{c}} & = e^{-\frac{1}{c} \ln{[ax^{-c}+by^{-c}]}} \\
& = e^{-\frac{1}{c} \ln{[ae^{-c\ln(x)}+be^{-c \ln(y)}]}} \\
& = e^{-\frac{1}{c} \ln{[a(1-c \ln(x) + o(c)) + b(1-c \ln(y) + o(c))}]} \\
& = e^{-\frac{1}{c} \ln{[a+b - c (a \ln(x)+b \ln(y)) + o(c))]}} \\
\end{align\*}
If $a+b>1$, the term in the exponential converges to $-\infty$ when $c \rightarrow 0$ and therefore $U(x,y) \rightarrow 0$. If $a+b<1$, the term converges to $+\infty$ and therefore $U(x,y) \rightarrow +\infty$.

To obtain the convergence towards the Cobb-Douglas function, we must therefore assume $a+b=1$. In that case we have
\begin{align\*}
U(x,y) & = e^{-\frac{1}{c} \ln{[1-c(a \ln(x) + (1-a) \ln(y))+o(c)]}} \\
& = e^{-\frac{1}{c} [-c(a \ln(x) + (1-a) \ln(y))+o(c)]} \\
& = e^{a \ln(x) + (1-a) \ln(y) + \frac{o(c)}{c}} \\
& \rightarrow\_{c \rightarrow 0} e^{a \ln(x) + (1-a) \ln(y)} \\
& = x^{a} y^{1-a}
\end{align\*}
which is the Cobb-Douglas utility function with parameters $(a,1-a)$.

> **Comments:**
> **username_0**: How did you get rid of the error term o(c)/c?  
> **username_3**: @username_0 By definition of $o(c)$, the fraction $\frac{o(c)}{c}$ converges to $0$ when $c$ tends to $0$.  

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