\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!

  • 1
    $\begingroup$ 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. $\endgroup$
    – Giskard
    Feb 3, 2016 at 9:12
  • $\begingroup$ @denesp Sorry was mistake, I corrected it. Was supposed to be 1. $\endgroup$
    – bloopton
    Feb 3, 2016 at 14:48
  • $\begingroup$ @AlecosPapadopoulos I found that question and that's what confused me. I'm studying consumer theory right now, not producer theory. $\endgroup$
    – bloopton
    Feb 3, 2016 at 14:49
  • 2
    $\begingroup$ 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. $\endgroup$ Feb 3, 2016 at 17:36

1 Answer 1


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 $(a,1-a)$.

  • $\begingroup$ How did you get rid of the error term o(c)/c? $\endgroup$
    – bloopton
    Feb 4, 2016 at 6:04
  • 1
    $\begingroup$ @bloopton By definition of $o(c)$, the fraction $\frac{o(c)}{c}$ converges to $0$ when $c$ tends to $0$. $\endgroup$
    – Oliv
    Feb 4, 2016 at 7:15

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