<blockquote>
<p>But how can I derive from first the second equation? <span class="math-container">$lim_{t
 \rightarrow \infty} \beta^t U'(f(k_t)-k_{t+1})k_{t+1}=lim_{t
 \rightarrow \infty} \beta^{t-1} U'(f(k_{t-1})-k_{t})k_{t}$</span></p>
</blockquote>
<p>You can notice that the second side of the equation is just a relabeling of the first side, where  <span class="math-container">$t$</span> is replaced by  <span class="math-container">$t-1$</span>, so that they actually are the same equation.</p>
<p>In the second side of the equation we have <span class="math-container">$lim_{t
 \rightarrow \infty}$</span>, even if, re-labeling, one could think that it should be   <span class="math-container">$lim_{(t-1)
 \rightarrow \infty}$</span>.  But it is the same, as <span class="math-container">$(t-1) \rightarrow \infty \Leftrightarrow t \rightarrow \infty$</span> (<span class="math-container">$-1$</span> is a fixed number, so <span class="math-container">$t$</span> must go to infinity).</p>

CC BY-SA 4.0
2023-03-18T21:03:25.233Z
2023-03-18T21:03:25.233Z
BakerStreet
<p>In Kruger's note, the process of deriving TVC is expressed as:<br />
<span class="math-container">$0=lim_{t \rightarrow \infty} \beta^t U'(f(k_t)-k_{t+1})k_{t+1}=lim_{t \rightarrow \infty} \beta^{t-1} U'(f(k_{t-1})-k_{t})k_{t}
=lim_{t \rightarrow \infty} \beta^{t-1}\beta U'(f(k_{t})-k_{t+1})f'(k_t)k_{t}=lim_{t \rightarrow \infty} \beta^{t} U'(f(k_{t})-k_{t+1})f'(k_t)k_{t}$</span><br />
The third equation is from Euler equation.<br />
But how can I derive from first the second equation?<br />
<span class="math-container">$lim_{t \rightarrow \infty} \beta^t U'(f(k_t)-k_{t+1})k_{t+1}=lim_{t \rightarrow \infty} \beta^{t-1} U'(f(k_{t-1})-k_{t})k_{t}$</span></p>

<macroeconomics>
Process of deriving TVC(transversality conditon)
guest
# Process of deriving TVC(transversality conditon)

(Asked by: username_0 on 2023-03-18)

In Kruger's note, the process of deriving TVC is expressed as:  
$0=lim\_{t \rightarrow \infty} \beta^t U'(f(k\_t)-k\_{t+1})k\_{t+1}=lim\_{t \rightarrow \infty} \beta^{t-1} U'(f(k\_{t-1})-k\_{t})k\_{t}
=lim\_{t \rightarrow \infty} \beta^{t-1}\beta U'(f(k\_{t})-k\_{t+1})f'(k\_t)k\_{t}=lim\_{t \rightarrow \infty} \beta^{t} U'(f(k\_{t})-k\_{t+1})f'(k\_t)k\_{t}$  
The third equation is from Euler equation.  
But how can I derive from first the second equation?  
$lim\_{t \rightarrow \infty} \beta^t U'(f(k\_t)-k\_{t+1})k\_{t+1}=lim\_{t \rightarrow \infty} \beta^{t-1} U'(f(k\_{t-1})-k\_{t})k\_{t}$

**Tags:** macroeconomics


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## 1 answers

### Answer by username_1 on 2023-03-18. Score: 2 | Accepted answer

> But how can I derive from first the second equation? $lim\_{t
> \rightarrow \infty} \beta^t U'(f(k\_t)-k\_{t+1})k\_{t+1}=lim\_{t
> \rightarrow \infty} \beta^{t-1} U'(f(k\_{t-1})-k\_{t})k\_{t}$

You can notice that the second side of the equation is just a relabeling of the first side, where $t$ is replaced by $t-1$, so that they actually are the same equation.

In the second side of the equation we have $lim\_{t
\rightarrow \infty}$, even if, re-labeling, one could think that it should be $lim\_{(t-1)
\rightarrow \infty}$. But it is the same, as $(t-1) \rightarrow \infty \Leftrightarrow t \rightarrow \infty$ ($-1$ is a fixed number, so $t$ must go to infinity).


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