## FANDOM

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