TITLE: Dividing within limits
QUESTION [0 upvotes]: Assume $\phi \in (0,1)$ and the following condition holds
$$\lim_{t\to\infty} \phi^t x_t = 0.$$
Is it true that 
$$\lim_{t\to\infty} ( \phi x_t - x_{t-1}) = 0?$$
My proof would be like this:
From the limit condition above, it follows that 
$$\lim_{t\to\infty} \phi^{t-1} x_{t-1} = 0$$
as well as
$$\lim_{t\to\infty} \phi^{t^2} x_{t} = 0.$$
Hence, we have
$$\lim_{t\to\infty} \phi^t x_t  - \lim_{t\to\infty} \phi^{t-1} x_{t-1} = \lim_{t\to\infty} \phi^{t-1}( \phi x_t - x_{t-1}) = \lim_{t\to\infty} \phi^{t^2} x_{t} = 0.$$
Now, divide by $\phi^{t-1}$ and the proposition follows. The only thing I am not sure about is whether I am allowed to divide by $\phi^{t-1}$ or not.
Thank you for your help!

REPLY [1 votes]: In general, what you are doing does not work. What I mean by that is, if you have $\lim_{n \rightarrow \infty} y_n x_n = 0$, then you cannot "divide by" $y_n$ and conclude $\lim_{n \rightarrow \infty} x_n = 0$. 
A concrete example is $y_n := 1/n$ and $x_n := 1$. Here we have $\lim_{n \rightarrow} y_n x_n = \lim_{n \rightarrow \infty} 1/n = 0$, but $\lim_{n \rightarrow \infty} x_n = 1$. 
As a matter of fact the claim $\lim_{t \rightarrow \infty} (\phi x_t - x_{t-1}) = 0$ is alltogether false. A counterexample is $\phi = \frac{1}{2}$ and $x_n = 1$. Then we have $\phi \in (0,1)$ and $\lim_{t \rightarrow \infty} \phi^t x_t = \lim_{t \rightarrow \infty} \frac{1}{2^t} = 0$, but $\lim_{t \rightarrow \infty} (\phi x_t - x_{t-1}) = \lim_{t \rightarrow \infty} - \frac{1}{2} = - \frac{1}{2} \neq 0$.
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