Combination Theorem for Sequences/Quotient Rule

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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:

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


Then:

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

provided that $m \ne 0$.


Proof

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

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

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

Now, for $n > N$, consider:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \left\vert \frac {x_n} {y_n} - \frac {l} {m} \right\vert\) \(=\) \(\displaystyle \left\vert \frac {m x_n - y_n l} {m y_n} \right\vert\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(<\) \(\displaystyle \frac 2 {\left\vert{m}\right\vert^2} \left\vert{m x_n - y_n l}\right\vert\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    

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

The result follows by the Squeeze Theorem for Sequences of Complex Numbers (which applies as well to real as to complex sequences).

$\blacksquare$


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