The Algebraic Limit Theorem
Let \(\lim x_n = x\) and \(\lim y_n = y\). Then, \(\lim (x_n + y_n) = x + y\).

Strategy/Notes

We want to show that \(\lim (x_n + y_n) = x + y\). So we want to find \(N\) such that for any \(\epsilon > 0\),

$$ \begin{align*} \lvert x_n + y_n - (x + y) \rvert < \epsilon. \end{align*} $$

We can use the triangle inequality to bound the term:

$$ \begin{align*} \lvert x_n + y_n - (x + y) \rvert &= \lvert (x_n - x) + (y_n - y) \rvert \\ &\leq \lvert x_n - x \rvert + \lvert y_n - y \rvert. \end{align*} $$

So now the goal is to get the right hand side to be \(\epsilon\). To do so we’re given that \(\lim x_n = x\). This means that we can make \(\lvert x_n - x \rvert\) be as small as we want and similarly we can make \(\lvert y_n - y \rvert\) be as small as we want. So if we make each of them be \(\epsilon/2\) then the terms will total to \(\epsilon\) which is what we want. Formally, given that \(\lim x_n = a\), then we know that for some \(n \geq N_1\),

$$ \begin{align*} \lvert x_n - x \rvert < \frac{\epsilon}{2}. \end{align*} $$

Similarly, given that \(\lim y_n = y\), then we know that for some \(n \geq N_2\),

$$ \begin{align*} \lvert y_n - y \rvert < \frac{\epsilon}{2}. \end{align*} $$

You can see now if we replace these terms in the equation above we will get

$$ \begin{align*} \lvert (x_n - x) + (y_n - y) \rvert &\leq \lvert x_n - x \rvert + \lvert y_n - y \rvert = \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon. \end{align*} $$

\(N\) can safely be set to the maximum of \(N_1\) and \(N_2\) to guarantee the above bounds.

Formal Proof

Let \(\epsilon > 0\) be arbitrary. Since we want to bound the sum \(\lvert (x_n - x) + (y_n - y) \rvert\), we apply the definition of the limit to each sequence using \(\epsilon/2\). Since \(\lim x_n = x\), then there exists an \(N_1 \in \mathbb{N}\) such that for all \(n \geq N_1\), we must have

$$ \begin{align*} \lvert x_n - x \rvert < \frac{\epsilon}{2}. \end{align*} $$

Similarly, since \(\lim y_n = y\), then there exists an \(N_2 \in \mathbb{N}\) such that for all \(n \geq N_2\), we must have

$$ \begin{align*} \lvert y_n - y \rvert < \frac{\epsilon}{2}. \end{align*} $$

Therefore, choose \(N\) such that \(N = \max(N_1,N_2)\). Let \(n \geq N\). Then,

$$ \begin{align*} \lvert (x_n - x) + (y_n - y) \rvert &\leq \lvert x_n - x \rvert + \lvert y_n - y \rvert \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align*} $$

Thus, \(\lvert (x_n - x) + (y_n - y) \rvert < \epsilon\) for all \(n \geq N\) so by definition

$$ \begin{align*} \lim (x_n + y_n) = x + y. \end{align*} $$

as we wanted to show. \(\blacksquare\)

References