Combination Theorem for Sequences/Normed Division Ring/Inverse Rule/Lemma

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

Let $\sequence {y_n}$ be a sequence in $R$.

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


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

Suppose $l \ne 0$, and $\forall n: y_n \ne 0$.

Then:
 * $\displaystyle \lim_{n \mathop \to \infty} y_n^{-1} = l^{-1}$.

Proof
Let $\epsilon > 0$ be given.

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

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

Since $\sequence {y_n}$ converges to $l \ne 0$, by Sequence Converges to Within Half Limit then:
 * $\exists N_2 \in \N: \forall n \gt \N_2: \dfrac {\norm l} 2 \lt \norm {y_n}$.

or equivalently:
 * $\exists N_2 \in \N: \forall n \gt \N_2: 1 \lt \dfrac {2 \norm {y_n} } {\norm l }$.

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

Thus $\forall n > N$:
 * $1 \lt \dfrac {2 \norm {y_n} } {\norm l }$.
 * $ \norm {y_n - l} < \epsilon'$.

Hence:

Hence:
 * $\sequence { {y_n}^{-1} }$ is convergent with $\displaystyle \lim_{n \mathop \to \infty} {y_n}^{-1} = l^{-1}$.