Combination Theorem for Sequences/Normed Division Ring/Quotient Rule

Theorem
Let $\struct {R, \norm {\, \cdot \,} }$ be a commutative normed division ring with zero: $0$.

That is, $\struct {R, \norm {\, \cdot \,} }$ is a valued field.

Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $R$.

Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limits:


 * $\displaystyle \lim_{n \mathop \to \infty} x_n = l$
 * $\displaystyle \lim_{n \mathop \to \infty} y_n = m$

Suppose $m \ne 0$.

Then:
 * $\exists K \in \N : \forall n > K : y_n \ne 0$.

and the sequence
 * $\sequence { \dfrac {x_{K+n}} {y_{K+n}} }_{n \in \N}$ is well-defined and convergent with $\displaystyle \lim_{n \mathop \to \infty} \dfrac {x_{K+n}} {y_{K+n}} = \dfrac l m$.

Proof
Since $\sequence {y_n}$ converges to $m \ne 0$, by Sequence Converges to Within Half Limit then:
 * $\exists K \in \N: \forall n \gt K: \dfrac {\norm m} 2 \lt \norm {y_n}$.

or equivalently:
 * $\exists K \in \N: \forall n \gt K: 0 \lt \dfrac 1 {\norm {y_n}} \lt \dfrac 2 {\norm m}$.

By Axiom (N1) of norm (Positive definiteness) $\forall n > K : y_n \ne 0$.

Let $\sequence {x'_n}$ be the subsequence of $\sequence {x_n}$ where $x'_n = x_{K+n}$.

By Limit of Subsequence equals Limit of Sequence then $\sequence {x'_n}$ is convergent and $\displaystyle \lim_{n \mathop \to \infty} x'_n = l$.

Similarly, let $\sequence {y'_n}$ be the subsequence of $\sequence {y_n}$ where $y'_n = y_{K+n}$.

Then $\sequence {y'_n}$ is convergent and $\displaystyle \lim_{n \mathop \to \infty} y'_n = m$.

It follows that $\sequence { \dfrac {x'_n} {y'_n} } $ is well-defined and $\sequence { \dfrac {x'_n} {y'_n} } = \sequence { \dfrac {x_{K+n}} {y_{K+n}} }_{n \in \N}$.

We now show that $\sequence { \dfrac {x'_n} {y'_n} }$ is convergent with $\displaystyle \lim_{n \mathop \to \infty} \dfrac {x'_n} {y'_n} = \dfrac l m$.

Let $M = \max \set {\norm l, \norm m}$.

Let $\epsilon > 0$ be given.

Let $\epsilon' = \dfrac {\epsilon {\norm m}^2} {4M}$, then $ \epsilon' > 0$.

Since $\sequence {x'_n} \to l$, as $n \to \infty$, we can find $N_1$ such that:
 * $\forall n > N_1: \norm {x'_n - l} < \epsilon'$

Similarly, for $\sequence {y'_n}$, we can find $N_2$ such that:
 * $\forall n > N_2: \norm {y'_n - m} < \epsilon'$

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

Thus $\forall n > N$:
 * $ 0 \lt \dfrac 1 {\norm {y'_n}} \lt \dfrac 2 {\norm m}$.
 * $ \norm {x'_n - l} < \epsilon'$.
 * $ \norm {y'_n - m} < \epsilon'$.

Hence:

Hence:
 * $\norm { \dfrac {x'_n} {y'_n} }$ is convergent with $\displaystyle \lim_{n \mathop \to \infty} \dfrac {x'_n} {y'_n} = \dfrac l m$.