<p>Consider the CRS production function
<span class="math-container">$$
F(K,L) = L f\left(\frac{K}{L}\right) \tag{1}
$$</span>
Let us write <span class="math-container">$F_1 \equiv \frac{\partial F}{\partial K}$</span> and <span class="math-container">$F_2 = \frac{\partial F}{\partial L}$</span>. Assume that:
<span class="math-container">$$
\begin{align*}
&amp;\lim_{K \to 0} F_1(K,L) = \infty,\tag{a}\\
&amp;\lim_{K \to \infty} F_1(K,L) = 0, \tag{b}\\
&amp;\lim_{L \to 0} F_2(K,L) = \infty, \tag{c}\\
&amp;\lim_{L \to \infty} F_2(K,L) = 0 \tag{d}
\end{align*}
$$</span></p>
<p>Notice that by differentiating <span class="math-container">$(1)$</span> with respect to <span class="math-container">$K$</span> and <span class="math-container">$L$</span> and defining <span class="math-container">$k = \frac{K}{L}$</span> we obtain:
<span class="math-container">$$
\begin{align*}
&amp;F_1(K,L) = L f'\left(\frac{K}{L}\right) \frac{1}{L} = f'(k), \tag{A}\\
&amp;F_2(K,L) = f\left(\frac{K}{L}\right) + L f'\left(\frac{K}{L}\right) \left(-\frac{K}{L^2}\right) = f(k) - k f'(k) \tag{B}
\end{align*}
$$</span></p>
<ol>
<li><p>Now, consider an arbitrary sequence <span class="math-container">$(k_n)_{n \in \mathbb{N}}$</span>.  Define sequences <span class="math-container">$(K_n)_{n \in \mathbb{N}}$</span> and <span class="math-container">$(L_n)_{n \in \mathbb{N}}$</span> such that <span class="math-container">$K_n = k_n$</span> and <span class="math-container">$L_n = 1$</span>. Notice that by definition <span class="math-container">$\frac{K_n}{L_n} = k_n$</span>. Then if <span class="math-container">$k_n \to 0$</span>​ we have by <span class="math-container">$(A)$</span> and <span class="math-container">$(a)$</span>:
<span class="math-container">$$
\lim_{n \to \infty} f'(k_n) = \lim_{n \to \infty} F_1(K_n, 1) = \lim_{K \to 0} F_1(K,1) = \infty.
$$</span>
As <span class="math-container">$k_n \to 0$</span> was arbitrary, we have
<span class="math-container">$$
\lim_{k \to 0} f'(k) = \infty.
$$</span>
Similarly if <span class="math-container">$k_n \to \infty$</span>, we have by <span class="math-container">$(A)$</span> and <span class="math-container">$(b)$</span>:
<span class="math-container">$$
\lim_{n \to \infty} f'(k_n) = \lim_{n \to \infty} F_1(K_n, 1) = \lim_{K \to \infty} F_1(K,1) = 0.
$$</span>
This gives that
<span class="math-container">$$
\lim_{k \to \infty} f'(k) = 0.
$$</span></p>
</li>
<li><p>So far, we haven't used <span class="math-container">$(B)$</span> nor <span class="math-container">$(c)$</span> and <span class="math-container">$(d)$</span>. These imply another set of limiting conditions on <span class="math-container">$f'(.)$</span>. Indeed, let us define sequences <span class="math-container">$(K_n)_{n \in \mathbb{N}}$</span> and <span class="math-container">$(L_n)_{n \in \mathbb{N}}$</span>​ such that <span class="math-container">$K_n = 1$</span> and <span class="math-container">$L_n = \dfrac{1}{k_n}$</span>. Again we have that <span class="math-container">$\frac{K_n}{L_n} = k_n$</span>. Then for <span class="math-container">$k_n \to 0$</span> we have using <span class="math-container">$(B)$</span> and <span class="math-container">$(d)$</span>:
<span class="math-container">$$
\lim_{n \to \infty} (f(k_n) - k_n f'(k_n)) = \lim_{n \to \infty} F_2(1,L_n) = \lim_{L_n \to \infty} F_2(1,L_n) = 0
$$</span>
This gives that:
<span class="math-container">$$
\lim_{k \to 0} f(k_n) = \lim_{k \to 0} k f'(k).
$$</span>
Observe that if <span class="math-container">$f$</span> is continuous at <span class="math-container">$0$</span> then this implies that:
<span class="math-container">$$
f(0) = \lim_{k \to 0} k f'(k).
$$</span>
Next taking a sequence <span class="math-container">$k_n \to \infty$</span>, we have using <span class="math-container">$(B)$</span> and <span class="math-container">$(c)$</span> that:
<span class="math-container">$$
\lim_{n \to \infty} (f(k_n) - k_n f'(k_n)) = \lim_{n \to \infty} F_2(1, L_n) = \lim_{L \to 0} F_2(1,L) = \infty
$$</span>
As such,
<span class="math-container">$$
\lim_{k \to \infty} (f(k) - k f'(k)) = \infty)
$$</span></p>
</li>
</ol>

CC BY-SA 4.0
2021-08-24T08:21:46.997Z
2021-08-24T09:12:17.620Z
2021-08-24T09:12:17.620Z
tdm
<p>I want to show that the Inada conditions <span class="math-container">$\underset{K\rightarrow0}{\lim} \frac{\partial F}{\partial K}=\underset{L\rightarrow0}{\lim} \frac{\partial F}{\partial L}=\infty$</span> and <span class="math-container">$\underset{K\rightarrow\infty}{\lim} \frac{\partial F}{\partial K}=\underset{L\rightarrow\infty}{\lim} \frac{\partial F}{\partial L}=0$</span> imply that <span class="math-container">$\underset{k\rightarrow0}{\lim} f'(k)=\infty$</span> and <span class="math-container">$\underset{k\rightarrow\infty}{\lim} f'(k)=0$</span>.</p>
<p>Here, the production function is given by <span class="math-container">$F(K,AL)=ALf(k)$</span>. We assume that <span class="math-container">$f'(k)&gt;0$</span> and <span class="math-container">$f''(k)&lt;0$</span>.</p>
<p>I start with the observation that if <span class="math-container">$\underset{k\rightarrow0}{\lim} f'(k)=\infty$</span>, then <span class="math-container">$\underset{K\rightarrow0}{\lim} f'(k)=\infty$</span> and <span class="math-container">$\underset{L\rightarrow\infty}{\lim} f'(k)=\infty$</span>. The first condition is fulfilled by the first Inada condition. I am also able to derive that <span class="math-container">$f'(k)=\frac{1}{K}F(K,AL)-\frac{L}{K} \frac{\partial F}{\partial L}$</span>. How can I show that the second condition is true as well?</p>

<economic-growth>
Inada Conditions for Intensive Form of Production Function
kékszajkók
# Inada Conditions for Intensive Form of Production Function

(Asked by: username_0 on 2021-08-22)

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?

**Tags:** economic-growth


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## 1 answers

### Answer by username_1 on 2021-08-24. Score: 3

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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