limit of quotient is quotient of limits

Prove the following:

Let $X$ be a metric space. Let $E \subseteq X$. Let $p$ be a limit point of $E$, and let $f,g: E \subseteq X \to \mathbb{K}$ be a maps. Suppose $\lim_{x \to p} f(x) = q_1, \lim_{x \to p} g(x) = q_2$. Then, if we define

$$\frac{f}{g}: Z= \{x \in E \mid g(x) \neq 0\} \to \mathbb{K}: x \mapsto f(x)/g(x)$$

we have $\lim_{x \to p} \left(\frac{f}{g}\right)(x) = \frac{q_1}{q_2}$ provided that $q_2 \neq 0$.

In my proof, I use;

$\lim_{x \to p} f(x) = q$ iff for every sequence $(p_n)$ in $E \setminus \{p\}$ for which $p_n \to p$, we have $f(p_n) \to q$

Is my proof correct?

Proof:

First, we check that $p$ is a limit point of $Z$, so that talking about limits makes sense.

Since $\lim_{x \to p} g(x) = q_2 \neq 0$, there is $\delta > 0$ such that $|g(x) - q_2| < |q_2| \neq 0$ for all $x \in E$ satisfying $0 <d_X(x,p) < \delta.$ It follows that $g(x) \neq 0$ for $x \in (E\setminus \{p\}) \cap B_X(p, \delta)$.

Let $\epsilon > 0$. Then we know that we can pick $z \in B_X(p, \epsilon \land \delta) \cap (E \setminus \{p\})$, since $p$ is a limit point of $E$. It follows that $z \in B_X(p,\epsilon) \cap (Z \setminus \{p\})$, and $p$ is a limit point of $Z$.

Notice that $\lim_{x \to p, x \in Z} f(x) = q_1$ and $\lim_{x \to p, x \in Z} g(x) = q_2$ (this follows trivially from the definition of limit).

Let $(p_n)_n$ be a sequence in $Z \setminus \{p\}$. Then, we know that $f(p_n) \to q_1$ and $g(p_n) \to q_2$. Hence, $$\frac{f(p_n)}{g(p_n)} \to \frac{q_1}{q_2}$$

and the result follows $\square$.

It looks pretty solid, but more justification may be needed at the end. What is $\Bbb K$? Have you got a previous result to rely on that says, if $(a_n)_n$ and $(b_n)_n$ are sequences in $\Bbb K$ with $a_n\to a\in\Bbb K$ and $b_n\to b\in\Bbb K,$ such that $b$ and each $b_n$ are non-zero, then $\frac{a_n}{b_n}\to\frac{a}{b}$? If so, then I'd say you're good to go. If not, then you'll have to prove this, first, either as a Lemma or as part of your proof.

Also, in this line, there's an issue:

Since $\lim_{x \to p} g(x) = q_2 \neq 0$, there is $\delta > 0$ such that $|g(x) - q_2| < |q_2| \neq 0$ for all $x \in E$ satisfying $0 <d_X(x,p) < \delta.$

While this is true, and while it does follow that $g$ is non-zero for $x$ sufficiently close to $p,$ you haven't actually shown it to be true. What specific $\epsilon>0$ could you use to find such an appropriate $\delta$?

• Ah sorry, I forgot to mention. $\mathbb{K}$ is the complex numbers or the real numbers. But I think any vector space will do. – user370967 Aug 1 '18 at 12:13
• What exactly is the issue? I clearly pick $\epsilon = |q_2| > 0$. – user370967 Aug 1 '18 at 12:17
• Ah! I see. The "$|q_2|\neq 0$" part was throwing me. Perhaps it would be clearer to say something like "Since $q_2\neq 0,$ then $|q_2|>0,$ so there is $\delta>0$ such that $|g(x)-q_2|<|q_2|$ by definition of limit." – Cameron Buie Aug 1 '18 at 12:21
• You are right. That would have been clearer. Thanks for your time! – user370967 Aug 1 '18 at 12:22
• As for $\Bbb K$ being an arbitrary vector space: division isn't necessarily defined in a given vector space, so I doubt it. It would have to be a field, at least. – Cameron Buie Aug 1 '18 at 12:22