If $x_n\to 1$ can we compute this indeterminate limit? If we let $x_n\to1$ and $y_n\to1$ can we understand the convergence of $$\dfrac{1-2x_n + x_n^2y_n}{(1-x_n)^2}?$$
This is a problem I've been thinking about for a while but can't actually pin down a solution. It feels like this should converge to $1$ as well but I fear the rate of convergence of $y_n$ makes a difference.
Could we say that the numerator is equal to $(1-x_n)^2(1+o(1))$ and so the limit is just $1$? I fear this may be hiding some issues inside the little o and hence not actually rigorous. 
 A: The expression can be written as $$1 + x_n^2 \frac {y_n -1}{(1-x_n)^2}$$
So the behavior entirely depends on how quickly $y_n\rightarrow 1$ compared to $x_n\rightarrow 1$. It may converge or may diverge.
Here are a couple of examples.
Example 1: if
$x_n=1+\frac 1 n $ and $y_n=1+\frac 1 n $ then clearly the whole thing behaves as $$\frac {\frac 1n} {\frac 1 {n^2}}=n$$ and diverges to $+\infty$.
Example 2: if
$x_n=1+\frac 1 {\sqrt n} $ and $y_n=1+\frac 1 n $ then the expression behaves as $$1+\frac {\frac 1n} {\frac 1 {n}}=1$$ and converges to 2$.
A: It doesn't converge to $1$ in general.
We have
$$\dfrac{1-2x_n + x_n^2y_n}{(1-x_n)^2} = 1 + \frac{x_n^2(y_n-1)}{(1-x_n)^2}$$
Consider $x_n = 1-\frac1{n^2}$ and $y_n = 1+\frac1n$. We have
$$\frac{x_n^2(y_n-1)}{(1-x_n)^2} = \frac{\left(1-\frac1{n^2}\right)^2\frac1n}{\frac1{n^4}} = n^3 \left(1-\frac1{n^2}\right)^2 \xrightarrow{n\to\infty} +\infty$$
A: We cannot determine the limit with the given information, indeed just consider for example $y_n=1/x_n\to 1$ we obtain $$\dfrac{1-2x_n + x_n^2y_n}{(1-x_n)^2}=\dfrac{1}{1-x_n}$$ and then


*

*for $x_n=1-\frac1n\to 1$ the limit is $\infty$

*for $x_n=1+\frac1n \to 1$ the limit is $-\infty$
