Algebra of limits

The algebra of limits is a set of rules for how limits may be manipulated with other operators.

For real sequences

For real convergent sequences $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ where $\lim_{n \to \infty} x_n = x$ and $\lim_{n \to \infty} y_n = y$, and for a real number $c\in\R$:

1. $\lim_{n \to \infty} x_n + y_n = x + y$

1. $\lim_{n \to \infty} c.x_n = c.x$

1. $\lim_{n \to \infty} x_n.y_n = x.y$

If also $y \ne 0$ and $\forall n \in \N : y_n \ne 0,$

4. $\lim_{n \to \infty} \frac {x_n}{y_n} = \frac{x}{y}$

Proof

1. Fix $\epsilon > 0.$
   Because $\lim_{n \to \infty} x_n = x,$ for $\epsilon^\prime = \frac{\epsilon}{2} \exists N_1 \in \N : \forall n \ge N_1, |x_n - x| < \frac{\epsilon}{2},$ and
   because $\lim_{n \to \infty} y_n = y,$ for $\epsilon^\prime = \frac{\epsilon}{2} \exists N_2 \in \N : \forall n \ge N_2, |y_n - y| < \frac{\epsilon}{2}.$
   Set $N = max\left\{{N_1, N_2}\right\}.$
   By the triangle inequality, $|(x_n + y_n) - (x + y)| = |(x_n - x) + (y_n - y)| \le |x_n - x| + |y_n - y|,$
   and $\forall n \ge N, |x_n - x| + |y_n - y| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$
   So $\forall n\geq N,|(x_{n}+y_{n})-(x+y)|<\epsilon ,$ ie $\lim_{n \to \infty} x_n + y_n = x + y.$

2. If $c = 0$, then $\forall n \in \N x_n = 0$, so it converges to 0. If $c \ne 0$, fix $\epsilon > 0.$
   Because $\lim_{n \to \infty} x_n = x,$ for $\epsilon^\prime = \frac{\epsilon}{|c|} \exists N \in \N : \forall n \ge N, |x_n - x| < \frac{\epsilon}{|c|},$
   So $|c.x_{n}-c.x|=|c||x_{n}-x|<|c|{\frac {\epsilon }{|c|}}=\epsilon ,$ ie $\lim_{n \to \infty} c.x_n = c.x$

3. Fix $\epsilon > 0.$ As $\left \langle {y_n} \right \rangle$ is convergent, it is bounded by some $M \in \R,$ ie $\forall n \in \N : |y_n| \le M.$
   Because $\lim_{n \to \infty} x_n = x,$ for $\epsilon^\prime = \frac{\epsilon}{2M} \exists N_1 \in \N : \forall n \ge N_1, |x_n - x| < \frac{\epsilon}{2M},$ and
   because $\lim_{n \to \infty} y_n = y,$ for $\epsilon^\prime = \frac{\epsilon}{2|x|+1} \exists N_2 \in \N : \forall n \ge N_2, |y_n - y| < \frac{\epsilon}{2|x|+1}.$
   Set $N = max\left\{{N_1, N_2}\right\}.$
   By the triangle inequality, and boundedness of $\left \langle {y_n} \right \rangle$, $|x_n.y_n - x.y| = |x_n.y_n - x.y_n + x.y_n - x.y| \le |y_n||x_n - x| + |x||y_n - y| \le M|x_n - x| + |x||y_n - y|$ And $\forall n \ge N, M|x_n - x| + |x||y_n - y| < M\frac{\epsilon}{2M} + |x|\frac{\epsilon}{2|x|+1} \le \frac{\epsilon}{2} + \frac{\epsilon}{2}.$
   So $\forall n \ge N, |x_n.y_n - x.y| < \epsilon,$ ie $\lim_{n \to \infty} x_n.y_n = x.y.$

4. By part 3, $\lim_{n \to \infty} \frac {x_n}{y_n} = (\lim_{n \to \infty} x_n)(\lim_{n \to \infty} \frac {1}{y_n}) = x\lim_{n \to \infty} \frac {1}{y_n}.$
   Fix $\epsilon > 0.$
   Because $\lim _{n\to \infty }y_{n}=y\neq 0,$ for $\epsilon ^{\prime }={\frac {|y|}{2}}\exists N_{1}\in \mathbb {N} :\forall n\geq N_{1},|y_{n}-y|<{\frac {|y|}{2}}.$
   Also, for $\epsilon^\prime = \frac{y^2\epsilon}{2} \exists N_2 \in \N : \forall n \ge N_2, |y_n - y| < \frac{y^2\epsilon}{2}.$
   By the triangle inequality, $|y| = |y_n + y - y_n| \le |y_n| + |y - y_n| \iff |y| - |y - y_n| \le |y_n|.$
   And for $n \ge N_1, |y_n - y| < \frac{|y|}{2} \iff |y| - |y_n - y| > \frac{|y|}{2},$ so $|y_n| \ge |y| - |y_n - y| > \frac{|y|}{2} \iff \frac{1}{|y_n|} < \frac{2}{|y|}.$
   Now $|\frac{1}{y_n} - \frac{1}{y}| = |\frac{y - y_n}{y_n.y}| = \frac{1}{|y_n||y|}|y - y_n| < \frac{2}{|y||y|}|y - y_n| = \frac{2}{y^2}|y - y_n|.$
   And now $\forall n \ge N_2, |y_n - y| < \frac{y^2\epsilon}{2} \iff \frac{2}{y^2}|y - y_n| < \epsilon.$
   Setting $N = max\left\{{N_1, N_2}\right\},$ we have $\forall n \ge N : |\frac{1}{y_n} - \frac{1}{y}| < \frac{2}{y^2}|y - y_n| < \epsilon,$
   or in other words $\lim_{n \to \infty} \frac {1}{y_n} = y.$
   Therefore, $\lim_{n \to \infty} \frac {x_n}{y_n} = x\lim_{n \to \infty} \frac {1}{y_n} = \frac{x}{y}.$