A result in asymptotic theory says the following: Let $Y$ be a real random variable with full support and $E(|Y|)<\infty$. Assume that: 
$$
(A)\quad \lim_{y\rightarrow \infty} \frac{\Pr(Y\geq y)}{\Pr(Y\geq \kappa h(y))}=1
$$ 
for some $\kappa>0$ and function $h$ such that $\lim_{y\rightarrow \infty} h'(y)<\infty$. Then, $\kappa=1$. 

Could you help me understand:**(1)** How to interpret $(A)$; **(2)** How to interpret the result that $\kappa$ must be 1; **(3)** Why would you expect this result to intuitively hold or not hold?

_________

***Thoughts.*** Regarding **(1)**: Suppose that $ h $ is ***increasing*** in $y $. Then,
$$
(B)\quad \lim_{y \rightarrow \infty} \Pr(Y \geq y) = \lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y)) = 0.
$$
In this setting with $h$ increasing, Assumption $(A)$ is stronger than $(B)$, because I think  $(A)$ says that $\Pr(Y \geq y)$ and $\Pr(Y \geq \kappa h(y))$ go to zero at the same rate (but I am not sure here about this). If I am correct, what is an example of function $h$ such that $(B)$ holds but $(A)$ does not?


When $h$ is ***not*** increasing, in general, we may find that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ does not exist. However, does statement (A) still imply that $\lim_{y \rightarrow \infty} \Pr(Y \geq \kappa h(y))$ exists and must be equal to zero?
 
Regarding **(2)**:   Somehow, this result suggests that if $(A)$ holds, then $h$ can be anything (can shift, rescale, or apply nonlinear transformations to $y$), but the multiplicative constant $\kappa$ in front of $h$ must be one. Is this a correct interpretation of the result?

Regarding **(3)**:  I have zero intuition. If my interpretation of the result is correct (see above), why can $h$ be anything and $\kappa$ must instead be one?