limit $\lim_\limits{x\to\infty} f(x)=x^a\left(1-\frac {0.1\prod_{t=1}^x(1-y(t))}{10}\right)^{x-a} $ How can we solve the following limit. Also, is f(x) monotonically increasing/decreasing.
$$\lim_\limits{x\to\infty} f(x)=x^a\left(1-\frac {0.1\prod_{t=1}^x(1-y(t))}{10}\right)^{x-a} $$
where $a\in Z^+$ is constant, and $x\in Z^+$. $0<y(t)<1$ and $\{y(t)\}$ is a none increasing sequence of numbers with the limit value of 0.
Also, if the behavior of f(x) depends on the sequence $\{y(t)\}$ , what conditions should $\{y(t)\}$  possess to guarantee convergence of f(x) to 0.
 A: Set
$$z_n=c\prod_{i=1}^n(1-y_i)\tag1$$
Here $c=0.01$. We're interested in the asymptotic behavior of
$$f(n)=n^a(1-z_n)^n(1-z_n)^{-a}$$
From the information we have about $(y_n)$, we can deduce that $(z_n)$ is a strictly decreasing positive sequence. Thus the factor $(1-z_n)^{-a}$ converges to some limit between $0$ and $1$, so for our purposes this is basically equivalent to the study of
$$f(n)=n^a(1-z_n)^n$$
In general not much can be said about such a sequence. Of course if $z_n$ does not converge to $1$, the geometric factor crushes the polynomial factor $n^a$ and $f(n)\to 0$. But when $z_n\to 1$, the behavior of $f(n)$ has to do with how fast that convergence is. If $z_n$ goes to $1$ quickly enough, then it "beats" the power of $n$ and $(1-z_n)^n$ does not converge to $0$, so $f(n)\to+\infty$. If $z_n$ goes to $1$ more slowly, then $(1-z_n)^n$ converges to $0$, and if this converges is itself sufficiently fast, it may dominate the polynomial factor and we'll have $f(n)$ bounded or maybe even $f(n)\to 0$.
Of course, we're not working in the fully general case: we have a formula for $z_n$. But how much information does this formula really give us? How does the information we have about $(y_n)$ restrict which sequences $(z_n)$ can be? We can express $y_n$ in terms of $z_n$ like so:
$$y_n=1-\frac{z_n}{z_{n-1}}\tag2$$
We know that $0<y_n<1$, that $(y_n)$ is non-increasing and converges to $0$, therefore the same is true of the right hand side of the equation. In fact, the two conditions (1) and (2) are equivalent if we take $c=1-z_1$. This means that as long as we choose $(z_n)$ so that the right hand side of (2) is non-increasing, contained in $(0, 1)$ and converging to $0$, then we can set $y_n$ according to (2) and we will have (1). This allows us to forget about the sequence $(y_n)$ and forget about infinite products, and focus on the (I think) simpler problem:

Consider
  $$f(n)=n^a(1-z_n)^n$$
  for $a$ a positive integer. If we know that the successive ratios $\frac{z_n}{z_{n-1}}$ are in $(0, 1)$, non-decreasing and converge to $1$, what can we say about the asymptotic behavior of $f(n)$?

The condition on the successive ratios of $(z_n)$ is basically a slowness condition: if $z_n$ converges to $1$, it does so "slowly", so we have reason to hope we can show $f(n)$ is bounded.
Unfortunately, it turns out that that hope is in vain. If we set $z_n=\frac1n$, it satisfies our slowness condition, and yet $f(n)\to+\infty$. 
We need more information about $(y_n)$.
