Is it possible for a sequence of random variable to converge to two different non-degenerate limit? Let's say we have a sequence of random variables $X_n$. Let $a_n^{(1)}, b_n^{(1)}, a_n^{(2)}, b_n^{(2)}$ be sequences of deterministic positive numbers.
Assuming that $b_n^{(1)}/b_n^{(2)} \to \infty$ as $n \to \infty$, is it possible that
$$
\frac{X_n - a_n^{(1)}}{b_n^{(1)}} \overset{d}{\to} Y_1
$$
and
$$
\frac{X_n - a_n^{(2)}}{b_n^{(2)}} \overset{d}{\to} Y_2
$$
such that $Y_1$ and $Y_2$ are two different non-degenerate random variables?
More precisely, is it possible that $Y_1$ and $Y_2$ are not constant a.s., and the total variation distance between them are not $0$.
 A: No.  This is the conclusion of Theorem 1 on p.40 of Gnedenko and Kolmogorov's Limit distributions, or of what Loeve calls a "Convergence of types theorem" (on p. 203 of his Probability Theory, 3d ed.)   After centering and scaling, all non-degenerate limit distributions of a sequence of distributions are of the same "type", that is, are related as $F$ and $G$ are when $F(x)=G(ax+b)$ for all $x$, for some finite $a\ne0$ and $b$.  The proof is not hard.  Everyone attibutes the result to Khinchin, but I don't have a precise reference.
A: Make each $X_i$ be uniform on $[0, 1]$. 
Let $b^{(1)}_n = n, b^{(2)}_n = 1$. 
Let $a^{(1)}_n = a^{(2)}_n = 0$. 
Then the first sequence converges to $X_1 = 0$, while the second converges to a  rv that's uniform on $[0, 1]$.
N.B.: It appears that you've used $X_1$ to denote both the first element of your sequence, and the first limit (and similarly for $X_2$). If that was intentional, which seems unlikely, then I need to modify my answer to this: 
"Make each $X_i$ be uniform on $[0, 1]$, except make $X_2$ be the everywhere-zero rv."
