asymptotic order in probability
Solution 1:
For any $\epsilon>0$, $$ \mathsf{P}(|X_n-1|\le \epsilon)=\mathsf{P}(1-\epsilon\le X_n\le 1+\epsilon)\to 1 $$ as $n\to\infty$, which implies that for any $M<1$, $$ \mathsf{P}(|X_n| \ge M)\ge \mathsf{P}(X_n \ge M)\to 1. $$