Combination Theorem for Sequences/Real

Theorem
Let $$\left \langle {x_n} \right \rangle$$ and $$\left \langle {y_n} \right \rangle$$ be sequences in $\mathbb{R}$.

Let $$\left \langle {x_n} \right \rangle$$ and $$\left \langle {y_n} \right \rangle$$ be convergent to the following limits:

$$\lim_{n \to \infty} x_n = l, \lim_{n \to \infty} y_n = m$$

Let $$\lambda, \mu \in \mathbb{R}$$ be any real numbers.

Then:


 * $$\lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m$$;


 * $$\lim_{n \to \infty} \left({x_n y_n}\right) = l m$$;


 * $$\lim_{n \to \infty} \frac {x_n} {y_n} = \frac l m$$, provided that $$m \ne 0$$.

Proof

 * $$\lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m$$:

First we show that $$\lim_{n \to \infty} \left({\lambda x_n}\right) = \lambda l$$.

Next we show that $$\lim_{n \to \infty} \left({x_n + y_n}\right) = l + m$$.

Let $$\epsilon > 0$$ be given. Then $$\frac \epsilon 2 > 0$$.

Since $$\lim_{n \to \infty} x_n = l$$, we can find $$N_1$$ such that $$\forall n > N_1: \left|{x_n - l}\right| < \frac \epsilon 2$$.

Similarly, since $$\lim_{n \to \infty} y_n = m$$, we can find $$N_2$$ such that $$\forall n > N_2: \left|{y_n - m}\right| < \frac \epsilon 2$$.

Now let $$N = \max \left\{{N_1, N_2}\right\}$$.

Then if $$n > N$$, both the above inequalities will be true.

Thus $$\forall n > N$$:

$$ $$ $$

Hence $$\lim_{n \to \infty} \left({x_n + y_n}\right) = l + m$$.


 * $$\lim_{n \to \infty} \left({x_n y_n}\right) = l m$$:


 * $$\lim_{n \to \infty} \frac {x_n} {y_n} = \frac l m$$, provided that $$m \ne 0$$: