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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, $


  1. $ \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 \ge N, |(x_n + y_n) - (x + y)| < \epsilon, $ ie $ \lim_{n \to \infty} x_n + y_n = x + y. $
  1. 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 $
  1. 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. $
  1. 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 \ne 0, $ for $ \epsilon^\prime = \frac{|y|}{2} \exists N_1 \in \N : \forall n \ge 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}. $
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