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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 $c\rightarrow 0$?

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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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