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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

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

  3. \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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