# Inada Conditions for Intensive Form of Production Function

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?

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, $$\lim_{k \to \infty} (f(k) - k f'(k)) = \infty)$$