Why can't we choose a simpler bound in this proof?

A basic property of limits is: "the limit of the product is the product of the limits". More precisely,

Claim: Let $$D \subseteq \mathbb{R}$$, $$a$$ be a cluster point of $$D$$, $$f: D \to \mathbb{R}$$, and $$g: D \to \mathbb{R}$$.

Suppose that $$\lim_{x \to a} f(x) = L$$ and $$\lim_{x \to a} g(x) = K$$. Then

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

Proof:

Since $$\lim_{x \to a} g(x) = K$$ and $$1 > 0$$, $$\exists \delta_1 > 0$$ such that $$\forall x \in D$$ with $$0 < |x-a| < \delta_1$$, $$|g(x) - K| < 1$$, whence $$|g(x)| < |K| + 1$$.

Since $$\lim_{x \to a} f(x) = L$$ and $$\frac{\epsilon}{2(|K| + 1)} > 0$$, $$\exists \delta_2 > 0$$ such that $$\forall x \in D$$ with $$0 < |x-a| < \delta_2$$, $$|f(x) - L| < \frac{\epsilon}{2(|K| + 1)}$$.

Since $$\lim_{x \to a} g(x) = K$$ and $$\frac{\epsilon}{2(|L| + 1)} > 0$$, $$\exists \delta_3 > 0$$ such that $$\forall x \in D$$ with $$0 < |x-a| < \delta_3$$, $$|g(x) - K| < \frac{\epsilon}{2(|L| + 1)}$$.

Let $$\delta = \min\{\delta_1, \delta_2, \delta_3\} > 0$$. Then, $$\forall x \in D$$ with $$0 < |x-a| < \delta$$,

$$|f(x)g(x) - LK| = |f(x)g(x) - Lg(x) + Lg(x) - LK|$$

$$\leq |g(x)||f(x) - L| + |L||g(x) - K|$$

$$\leq (|K| + 1)|f(x) - L| + |L||g(x) - K|$$ (since $$0 < |x-a| < \delta_1$$)

$$< (|K| + 1)\frac{\epsilon}{2(|K| + 1)} + |L||g(x) - K|$$ (since $$0 < |x-a| < \delta_2$$)

$$< \frac{\epsilon}{2} + |L| \frac{\epsilon}{2(|L| + 1)}$$ (since $$0 < |x-a| < \delta_3$$)

$$< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \Box$$

This is the proof as given in my lecture notes.

My question: In the third line, why do we choose to bound $$g$$ by $$\frac{\epsilon}{2(|L| + 1)}$$ instead of the slightly more simple $$\frac{\epsilon}{2|L|}$$? If we did this, the last two lines would instead be:

$$< \frac{\epsilon}{2} + |L| \frac{\epsilon}{2|L|}$$ (since $$0 < |x-a| < \delta_3$$)

$$= \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$.

We lose the last strong inequality, but we still ultimately get $$|f(x)g(x) - LK| < \epsilon$$, so does this work? I feel like there must be a reason for choosing the slightly more complicated bound.

• But what if $L=0$? Apr 5, 2020 at 16:07
• @BrianM.Scott Ah. So is that the only reason? Apr 5, 2020 at 21:37
• It appears to be, yes. Apr 5, 2020 at 21:42