0
$\begingroup$

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.

$\endgroup$
3
  • 3
    $\begingroup$ But what if $L=0$? $\endgroup$ Apr 5, 2020 at 16:07
  • $\begingroup$ @BrianM.Scott Ah. So is that the only reason? $\endgroup$ Apr 5, 2020 at 21:37
  • 1
    $\begingroup$ It appears to be, yes. $\endgroup$ Apr 5, 2020 at 21:42

0

You must log in to answer this question.

Browse other questions tagged .