I am trying to evaluate the limit of the CES Production function:

$$Y=(aK^p + bL^p)^{1/p} $$

when $p$ goes to -infinity.

It first yields the indeterminate form $0^0$. I tried solving the problem by taking the log of the function, but then the limit gave resulted in $0\cdot(-\infty)$.

Thanks in advance.


Let be $Y=(aK^p+bL^p)^{1/p}$ the CES function with $a+b=1$. Take the logarithm $$ \log Y=\frac{\log(aK^p+bL^p)}{p} $$ and use the L'Hopital's rule with $p$ as variable: $$ \lim_{p\to -\infty}\frac{\log(aK^p+bL^p)}{p}=\lim_{p\to -\infty}\frac{aK^p\log K+bL^{p}\log L}{aK^p+bL^p} $$ Dividing both the numerator and denominator by $M = \min(K,L)$ $$\lim_{p\to -\infty}\frac{aK^p\log K+bL^{p}\log L}{aK^p+bL^p}= \lim_{p\to -\infty}\frac{a(K/M)^p\log K+b(L/M)^{p}\log L}{a(K/M)^p+b(L/M)^p}=\log M $$ We get $(K/M)^p\to 0$ as $p\to -\infty$ for $K>M$ and $(L/M)^p\to 0$ as $p\to -\infty$ for $L>M$; and $K/M=1$ if $K=M$ or $L/M=1$ if $L=M$.

Subsequently, we obtain the corresponding Leontief function for $Y$ $$ \lim_{p\to -\infty}Y=M=\min(K,L). $$

  • $\begingroup$ According to you, $\frac{d}{dp} a K^p + b L^p = aK^p\log K+bL^{p}\log L$. $\endgroup$ – Michael Nov 18 '14 at 14:29
  • 1
    $\begingroup$ Yes, we have an exponential function of the type $f(p)=\alpha^p$ with first derivative $f'(p)=\alpha^p\log\alpha$. $\endgroup$ – alexjo Nov 18 '14 at 14:59
  • $\begingroup$ My bad. Not awake. $\endgroup$ – Michael Nov 18 '14 at 15:57

I think this is an easier way to solve it.

When $L<K$ we can factorize $L$ as follows:

$Y=\Big[bL^p \Big(\frac{aK^p}{bL^p}+1\Big)\Big]^{1/p}=b^{1/p} L \Big[\frac{a}{b}\big(\frac{K}{L}\big)^p +1\Big]^{1/p}=b^{1/p} L \Big[\frac{a}{b}\big(\frac{L}{K}\big)^{-p} +1\Big]^{1/p}$

$lim_{p \rightarrow - \infty} Y =L$ ,

since $b^{1/-\infty}=1$, and $lim_{p \rightarrow - \infty} \big(\frac{L}{K}\big)^{-p}=\big(\frac{L}{K}\big)^{\infty}=0$.

By symmetry, when $K<L$ (factoring $K$ instead of $L$), we obtain:

$lim_{p \rightarrow - \infty} Y =K$ .

  • $\begingroup$ My apologies, it was just a typo. As you may know (but not commented...), $b^{(1/-\infty)}=1$ and thus the limit is L. Therefore, the answer still holds. You may want to remove the negative vote from my answer. $\endgroup$ – Martin FS Dec 9 '14 at 1:48

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