# Process of deriving TVC(transversality conditon)

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?
$$lim_{t \rightarrow \infty} \beta^t U'(f(k_t)-k_{t+1})k_{t+1}=lim_{t \rightarrow \infty} \beta^{t-1} U'(f(k_{t-1})-k_{t})k_{t}$$

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).