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? \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!
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}} &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 $U(x,y) \rightarrow +\infty$.

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 $(a,1-a)$.