Is it valid to reference the limit product proof when proving the quotient proof?

Question

Let's assume we've proven that

$$\lim_{x\to a} f(x)g(x) = \lim_{x\to a}f(x) \lim_{x\to a}g(x)$$

Where $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist and are nonzero. If either are zero then we can reference a separate proof about multiplying against constants.

Anyhow, now we are faced with the notion of proving

$$\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$$

Is it allowable to basically say "Same as the product proof except we define $h(x) = \frac{1}{g(x)}$ and use that instead?

Answer 1

The correct statement is something like

Suppose that $\lim_{x \to a} f(x) = L$ and $\lim_{x\to a} g(x) = M \ne 0$, where both $L$ and $M$ are finite real numbers. Then $$ \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{L}{M}. $$

It is absolutely appropriate to use previously proved theorems in order to prove new theorems. In this case, the result that you give is one important ingredient, i.e. we can write $$ \frac{f(x)}{g(x)} = f(x) \cdot \frac{1}{g(x)}, $$ then use that the limit of a product is the product of the limits in order to conclude that $$ \lim_{x\to a} \frac{f(x)}{g(x)} = L \cdot \lim_{x \to a} \frac{1}{g(x)}. $$ This does leave one detail unresolved: do you know how to show that $$ \lim_{x\to a} \frac{1}{g(x)} = \frac{1}{M}? $$ Somehow, you need to prove that this limit works out the way you want. One way to do this is as follows:

Proof: Fix $\varepsilon > 0$. Since $\lim_{x\to a} g(x) = M$ and $2|M|^2 > 0$, we can use the definition of the limit to choose $\delta_1>0$ so that $$|M-g(x)| < 2|M|^2\varepsilon\tag{1}$$ for all $x$ with $|x-a|<\delta_1$. By similar reasoning, we may also choose $\delta_2>0$ so that $$ |g(x) - M| < |M| \implies -|M| < g(x) - M < |M| \implies |g(x)| < 2|M| \tag{2}$$ for all $x$ such that $|x-a| < \delta_2$. Fix $\delta = \min\{ \delta_1, \delta_2 \}$. If $|x-a| < \delta$, then both (1) and (2) must hold, and so $$ \left| \frac{1}{g(x)} - \frac{1}{M} \right| = \left| \frac{M - g(x)}{Mg(x)} \right| = \frac{ |M - g(x)|}{|M||g(x)|} < \frac{2|M|^2 \varepsilon}{|M| \cdot 2|M|} = \varepsilon.$$ Therefore $$ \lim_{x\to a} \frac{1}{g(x)} = \frac{1}{M}. $$

With this little lemma in hand, we now have the result, namely $$ \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} f(x) \cdot \lim_{x\to a} \frac{1}{g(x)} = \frac{L}{M}. $$ Note, however, that this intermediate step probably didn't save us much time or effort. We could prove this directly by examining the expression $$ \left| \frac{f(x)}{g(x)} - \frac{L}{M} \right| = \frac{ |Mf(x) - Lg(x)| }{|M||g(x)|}. $$ Bounding the numerator is slightly more complicated, but not by much.