Combination Theorem for Sequences/Real

Theorem
Let $$X$$ be one of the standard number fields $$\Q, \R, \C$$.

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

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 $$\lambda, \mu \in X$$.

Then the following results hold:

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

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

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

Proof of Sum of Limits
We need to show that $$\lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m$$.


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

Let $$\epsilon > 0$$.

We need to find $$N$$ such that $$\forall n > N: \left|{\lambda x_n - \lambda l}\right| < \epsilon$$.

If $$\lambda = 0$$ the result is trivial.

So, assume $$\lambda \ne 0$$.

Then $$\left|{\lambda}\right| > 0$$ from either:
 * The definition of absolute value if $$X = \Q$$ or $$X = \R$$;
 * The definition of complex modulus if $$X = \C$$.

Hence $$\frac {\epsilon} {\left|{\lambda}\right|} > 0$$.

We have that $$x_n \to l$$ as $$n \to \infty$$.

Thus it follows that $$\exists N: \forall n > N: \left|{x_n - l}\right| < \frac {\epsilon} {\left|{\lambda}\right|}$$.

That is, $$\forall n > N: \left|{\lambda}\right| \left|{x_n - l}\right| < \epsilon$$.

But we have:

$$ $$

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


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

Proof of Product of Limits
We need to show that $$\lim_{n \to \infty} \left({x_n y_n}\right) = l m$$.

Since $$\left \langle {x_n} \right \rangle$$ converges, it is bounded by Convergent Sequence is Bounded.

Suppose $$\left|{x_n}\right| \le K$$ for $$n = 1, 2, 3, \ldots$$.

Then:

$$ $$ $$ $$ $$

But $$x_n \to l$$ as $$n \to \infty$$.

So $$\left|{x_n - l}\right| \to 0$$ as $$n \to \infty$$ from Convergent Sequence Minus Limit.

Similarly $$\left|{y_n - m}\right| \to 0$$ as $$n \to \infty$$.

From the above result $$\lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m$$, $$z_n \to 0$$ as $$n \to \infty$$.

The result follows by the Squeeze Theorem.

Proof of Quotient of Limits
We need to show that $$\lim_{n \to \infty} \frac {x_n} {y_n} = \frac l m$$, provided that $$m \ne 0$$.

As $$y_n \to m$$ as $$n \to \infty$$, it follows from Absolute Value of Limit that $$\left|{y_n}\right| \to \left|{m}\right|$$ as $$n \to \infty$$.

As $$m \ne 0$$, it follows from the definition of absolute value that $$\left|{m}\right| > 0$$.

From Sequence Converges to Within Half Limit, we have $$\exists N: \forall n > N: \left|{y_n}\right| > \frac {\left|{m}\right|} 2$$.

Now, for $$n > N$$, consider:

$$ $$

By the above, $$m x_n - y_n l \to ml - ml = 0$$ as $$n \to \infty$$.

The result follows from the corollary to the Squeeze Theorem for Sequences.