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I want to show that the Inada conditions $\underset{K\rightarrow0}{\lim} \frac{\partial F}{\partial K}=\underset{L\rightarrow0}{\lim} \frac{\partial F}{\partial L}=\infty$ and $\underset{K\rightarrow\infty}{\lim} \frac{\partial F}{\partial K}=\underset{L\rightarrow\infty}{\lim} \frac{\partial F}{\partial L}=0$ imply that $\underset{k\rightarrow0}{\lim} f'(k)=\infty$ and $\underset{k\rightarrow\infty}{\lim} f'(k)=0$.

Here, the production function is given by $F(K,AL)=ALf(k)$. We assume that $f'(k)>0$ and $f''(k)<0$.

I start with the observation that if $\underset{k\rightarrow0}{\lim} f'(k)=\infty$, then $\underset{K\rightarrow0}{\lim} f'(k)=\infty$ and $\underset{L\rightarrow\infty}{\lim} f'(k)=\infty$. The first condition is fulfilled by the first Inada condition. I am also able to derive that $f'(k)=\frac{1}{K}F(K,AL)-\frac{L}{K} \frac{\partial F}{\partial L}$. How can I show that the second condition is true as well?

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Consider the CRS production function $$ F(K,L) = L f\left(\frac{K}{L}\right) \tag{1} $$ Let us write $F_1 \equiv \frac{\partial F}{\partial K}$ and $F_2 = \frac{\partial F}{\partial L}$. Assume that: $$ \begin{align*} &\lim_{K \to 0} F_1(K,L) = \infty,\tag{a}\\ &\lim_{K \to \infty} F_1(K,L) = 0, \tag{b}\\ &\lim_{L \to 0} F_2(K,L) = \infty, \tag{c}\\ &\lim_{L \to \infty} F_2(K,L) = 0 \tag{d} \end{align*} $$

Notice that by differentiating $(1)$ with respect to $K$ and $L$ and defining $k = \frac{K}{L}$ we obtain: $$ \begin{align*} &F_1(K,L) = L f'\left(\frac{K}{L}\right) \frac{1}{L} = f'(k), \tag{A}\\ &F_2(K,L) = f\left(\frac{K}{L}\right) + L f'\left(\frac{K}{L}\right) \left(-\frac{K}{L^2}\right) = f(k) - k f'(k) \tag{B} \end{align*} $$

  1. Now, consider an arbitrary sequence $(k_n)_{n \in \mathbb{N}}$. Define sequences $(K_n)_{n \in \mathbb{N}}$ and $(L_n)_{n \in \mathbb{N}}$ such that $K_n = k_n$ and $L_n = 1$. Notice that by definition $\frac{K_n}{L_n} = k_n$. Then if $k_n \to 0$​ we have by $(A)$ and $(a)$: $$ \lim_{n \to \infty} f'(k_n) = \lim_{n \to \infty} F_1(K_n, 1) = \lim_{K \to 0} F_1(K,1) = \infty. $$ As $k_n \to 0$ was arbitrary, we have $$ \lim_{k \to 0} f'(k) = \infty. $$ Similarly if $k_n \to \infty$, we have by $(A)$ and $(b)$: $$ \lim_{n \to \infty} f'(k_n) = \lim_{n \to \infty} F_1(K_n, 1) = \lim_{K \to \infty} F_1(K,1) = 0. $$ This gives that $$ \lim_{k \to \infty} f'(k) = 0. $$

  2. So far, we haven't used $(B)$ nor $(c)$ and $(d)$. These imply another set of limiting conditions on $f'(.)$. Indeed, let us define sequences $(K_n)_{n \in \mathbb{N}}$ and $(L_n)_{n \in \mathbb{N}}$​ such that $K_n = 1$ and $L_n = \dfrac{1}{k_n}$. Again we have that $\frac{K_n}{L_n} = k_n$. Then for $k_n \to 0$ we have using $(B)$ and $(d)$: $$ \lim_{n \to \infty} (f(k_n) - k_n f'(k_n)) = \lim_{n \to \infty} F_2(1,L_n) = \lim_{L_n \to \infty} F_2(1,L_n) = 0 $$ This gives that: $$ \lim_{k \to 0} f(k_n) = \lim_{k \to 0} k f'(k). $$ Observe that if $f$ is continuous at $0$ then this implies that: $$ f(0) = \lim_{k \to 0} k f'(k). $$ Next taking a sequence $k_n \to \infty$, we have using $(B)$ and $(c)$ that: $$ \lim_{n \to \infty} (f(k_n) - k_n f'(k_n)) = \lim_{n \to \infty} F_2(1, L_n) = \lim_{L \to 0} F_2(1,L) = \infty $$ As such, $$ \lim_{k \to \infty} (f(k) - k f'(k)) = \infty) $$

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