<p>The answer of @tdm shows you how it is done using De L'Hospital's rule. If you want to avoid this rule you can write <span class="math-container">$u(c)=\frac{c^{1-\theta}-1}{1-\theta}$</span> and note that for <span class="math-container">$\theta\ne 1$</span> this implies <span class="math-container">$u(1)=0$</span>. Differentiating with respect to <span class="math-container">$c$</span> then gives you <span class="math-container">$u'(c)=c^{-\theta}$</span> and therefore <span class="math-container">$\lim_{\theta \to 1}u'(c)=c^{-1}$</span>. Then integrate again to get <span class="math-container">$\lim_{\theta \to 1}u(c)=\ln(c)+K$</span> and comparing for <span class="math-container">$c=1$</span> gives you <span class="math-container">$K=0$</span>, so <span class="math-container">$\lim_{\theta \to 1}u(c)=\ln(c)$</span>.</p>

CC BY-SA 4.0
2021-04-29T08:24:41.647Z
2021-04-29T08:24:41.647Z
VARulle
<p>The derivative of <span class="math-container">$a^{x}$</span> with respect to <span class="math-container">$x$</span> is equal to <span class="math-container">$a^x \ln(a)$</span>.</p>
<p>As such, using the chain rule, the derivative of <span class="math-container">$c^{1-\theta}$</span> with respect to <span class="math-container">$\theta$</span> equals <span class="math-container">$c^{1-\theta} \ln(c) (-1)$</span></p>
<p>L'Hospital gives:</p>
<p><span class="math-container">$$
\lim_{\theta \to 1} \frac{c^{1-\theta}-1}{1 - \theta}\\
= \lim_{\theta \to 1} \frac{(c^{1-\theta}-1)'}{(1-\theta)'},\\
= \lim_{\theta \to 1} \frac{c^{1-\theta} \ln(c) (-1)}{(-1)},\\
= \frac{c^0 \ln(c)}{1} = \ln(c).  
$$</span></p>

CC BY-SA 4.0
2021-04-29T06:03:42.333Z
2021-04-29T06:03:42.333Z
tdm
<p>I have given the utility function for the Ramsey-Cass-Koopmans model, as follows:</p>
<p><span class="math-container">$$ u(c_t)=\frac{c^{1-\theta}-1}{1-\theta} $$</span></p>
<p>The claim is that as <span class="math-container">$\theta \rightarrow 1$</span>, then the funciton is logarithmic, that is</p>
<p><span class="math-container">$\lim_{\theta\to 1} u(c) = ln(c)$</span></p>
<p>I was told to use L'hôpital's rule, but I can't figure it out.</p>
<p>How can I prove this?</p>
<p>Thank you</p>

<macroeconomics>
Limit of utility function in Ramsey-Cass-Koopmans model
Darvid
# Limit of utility function in Ramsey-Cass-Koopmans model

(Asked by: username_0 on 2021-04-29)

I have given the utility function for the Ramsey-Cass-Koopmans model, as follows:

$$ u(c\_t)=\frac{c^{1-\theta}-1}{1-\theta} $$

The claim is that as $\theta \rightarrow 1$, then the funciton is logarithmic, that is

$\lim\_{\theta\to 1} u(c) = ln(c)$

I was told to use L'hôpital's rule, but I can't figure it out.

How can I prove this?

Thank you

**Tags:** macroeconomics

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## 2 answers

### Answer by username_2 on 2021-04-29. Score: 4 | Accepted answer

The derivative of $a^{x}$ with respect to $x$ is equal to $a^x \ln(a)$.

As such, using the chain rule, the derivative of $c^{1-\theta}$ with respect to $\theta$ equals $c^{1-\theta} \ln(c) (-1)$

L'Hospital gives:

$$
\lim\_{\theta \to 1} \frac{c^{1-\theta}-1}{1 - \theta}\\
= \lim\_{\theta \to 1} \frac{(c^{1-\theta}-1)'}{(1-\theta)'},\\
= \lim\_{\theta \to 1} \frac{c^{1-\theta} \ln(c) (-1)}{(-1)},\\
= \frac{c^0 \ln(c)}{1} = \ln(c).
$$


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### Answer by username_1 on 2021-04-29. Score: 5

The answer of @tdm shows you how it is done using De L'Hospital's rule. If you want to avoid this rule you can write $u(c)=\frac{c^{1-\theta}-1}{1-\theta}$ and note that for $\theta\ne 1$ this implies $u(1)=0$. Differentiating with respect to $c$ then gives you $u'(c)=c^{-\theta}$ and therefore $\lim\_{\theta \to 1}u'(c)=c^{-1}$. Then integrate again to get $\lim\_{\theta \to 1}u(c)=\ln(c)+K$ and comparing for $c=1$ gives you $K=0$, so $\lim\_{\theta \to 1}u(c)=\ln(c)$.


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