Problem: Suppose that $\left( x_n \right) $ and $\left( y_n \right) $ are two sequences with $x_n \to x$. Show that $y_n \to x$ if and only if $x_n - y_n \to 0$.
Solution: Let $y_n \to x$. Then it is trivial that $\left( x_n - y_n \right) $ converges to $0$ as both the sequence have the same limit. On the other hand, if $x_n - y_n \to 0$, then we need to show that $y_n \to x$. Let $\varepsilon >0$ be given. Since $x_n \to x$, then there exists $n_1\in \mathbb{N} $ such that for $n\geq n_1$ \[ \left\vert x_n - x \right\vert \lt \frac{\varepsilon}{2} . \] Similarly, there exists $n_2 \in \mathbb{N} $ such that for $n\geq n_2$ \[ \left\vert x_n - y_n \right\vert \lt \frac{\varepsilon }{2}. \] Therefore, for $n \geq \max\{ n_1, n_2 \} $, \begin{align*} \left\vert y_n - x \right\vert & = \left\vert y_n - x_n + x_n - x \right\vert \\ & \leq \left\vert y_n - x_n \right\vert + \left\vert x_n - x \right\vert \\ & \lt \frac{\varepsilon}{2} + \frac{\varepsilon }{2} = \varepsilon . \end{align*} Therefore, the sequence $\left( y_n \right) $ is converging to $x$.