2
$\begingroup$

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

Let $L_1 = \lim_{x \to a} f(x)$ and $L_2 = \lim_{x \to a} g(x)$. We assume $L_1, L_2 \neq 0$ (for scenarios where $L_1$ or $L_2$ are $0$, the result is trivially true).

Let $\epsilon > 0$. We need a $\delta$ such that $|f(x)g(x) - L_1L_2| < \epsilon$ whenever $0 < |x-a| < \delta$.

We can rewrite

$$\begin{align}|f(x)g(x) - L_1L_2| &= |(f(x)-L_1)(g(x)-L_2) + L_1(g(x)-L_2) + L_2(f(x)-L_1)| \\ &\leq |f(x)-L_1||g(x)-L_2| + |L_1||g(x)-L_2| + |L_2||f(x)-L_1| \\ &< \epsilon\end{align}$$

by triangle inequality.

Let $\delta_1 > 0$ such that $|f(x)-L_1| < \sqrt{\frac{\epsilon}{3}}$ whenever $0 < |x-a| < \delta_1$.

Let $\delta_2 > 0$ such that $|g(x)-L_2| < \sqrt{\frac{\epsilon}{3}}$ whenever $0 < |x-a| < \delta_2$.

Let $\delta_3 > 0$ such that $|g(x)-L_2| < \frac{\epsilon}{3|L_1|}$ whenever $0 < |x-a| < \delta_3$.

Let $\delta_4 > 0$ such that $|f(x)-L_1| < \frac{\epsilon}{3|L_2|}$ whenever $0 < |x-a| < \delta_4$.

Suppose that $0 < |x-a| < \delta$ where $\delta = \min(\delta_1, \delta_2, \delta_3, \delta_4)$. Then we have:

$$\begin{align}|f(x)-L_1||g(x)-L_2| &+ |L_1||g(x)-L_2| + |L_2||f(x)-L_1| \\&< \sqrt{\frac{\epsilon}{3}} \cdot \sqrt{\frac{\epsilon}{3}} + |L_1|\frac{\epsilon}{3|L_1|} + |L_2|\frac{\epsilon}{3|L_2|} \\&= \epsilon \end{align}$$

Is this a correct proof?

$\endgroup$
1
$\begingroup$

your proof of the essential point seems flawless. one might only raise one small quibble: the statement of what you are proving is a little imprecise. suppose $f(x)=x$ and $g(x)=\frac1x$. then $\forall x.f(x)g(x)=1$ so

$$ \lim_{x \to 0} f(x)g(x) = 1 $$ but what is $\lim_{x\to 0}f(x) \lim_{x \to 0}g(x)$?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm a little confused, does this make the entire proof wrong? Do I have to account for some extra case? What am I missing exactly? Accounting for infinity separately? Indeterminate forms? $\endgroup$ – user525966 Mar 1 '18 at 1:45
  • $\begingroup$ not at all, it is a very nice proof. you can try formulations like (abbreviated here) "if Limit f and Limit g exist, then Limit fg exists and Limit fg = Limit f * Limit g" $\endgroup$ – David Holden Mar 1 '18 at 1:49
  • $\begingroup$ your proof shows precisely that if the two limits exist then the product of the functions also has a limit and it is the product of the limits. but what you want to avoid is the implication that if the product of two functions has a limit then the functions separately each have limits. it is just a matter of careful phrasing. $\endgroup$ – David Holden Mar 1 '18 at 2:54
  • $\begingroup$ What do I need to add to make the proof more complete? $\endgroup$ – user525966 Mar 1 '18 at 2:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.