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I came up with this proof about Product theorem of limits after I watched "MIT Calculus Revisited" and I hope it's OK:

primal work on proof

Given $\hspace{0.5in}$ $$\mathop {\lim }\limits_{x \to a} f\left( x \right) = K \hspace{0.1in} and \hspace{0.1in} \mathop {\lim }\limits_{x \to a} g\left( x \right) = L$$

we hope to prove the existence of $$\mathop {\lim }\limits_{x \to a} [f\left( x \right) \cdot g\left( x \right)]= L \cdot K$$ means by Epsilon Delta definition :

$\hspace{1.4in}$every number $\varepsilon > 0$ there is some number $\delta > 0$ such that $$\left| {(f\left( x \right) \cdot g\left( x \right))-(L \cdot K) } \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta$$

$\hspace{.91in} \vert{(f\left( x \right) \cdot g\left( x \right))\vert-\vert(L \cdot K)\vert } < \varepsilon\hspace{.2in} \text{By Triangular inequality (1)} $

$\hspace{.91in} \vert{f\left( x \right)\vert \cdot \vert g\left( x \right)\vert- {\vert L \vert \cdot \vert K \vert}} < \varepsilon \hspace{0.2in} \text{by absoluate value product (2) }$

$\hspace{.91in} \vert{K + (f\left( x \right)-K) \vert \cdot \vert L + (g\left( x \right)-L)\vert- {\vert L \vert \cdot \vert K \vert}} < \varepsilon \hspace{0.2in}\text{Adding zeros)(3) }$

$\hspace{.91in} { \vert L \vert \cdot \vert(f\left( x \right)-K) \vert +\vert K \vert \cdot \vert(g\left( x \right)-L)\vert{}+\vert f\left( x \right)-K \cdot g\left( x \right)-L\vert } < \varepsilon \hspace{.08in}\text{simplify(4)}$

$\hspace{.9in}$put $\vert L \vert \cdot \vert f\left( x \right)-K\vert < \varepsilon/3$

$\hspace{.9in}$put $\vert K \vert \cdot \vert g\left( x \right)-L\vert < \varepsilon/3$

$\hspace{.9in}$put $\vert f\left( x \right)-K\vert < \sqrt{\varepsilon/3}$

$\hspace{.9in}$put $\vert g\left( x \right)-L\vert < \sqrt{\varepsilon/3}$

Actual proof

let $\varepsilon > 0 \hspace{0.1in}$ chooes $\varepsilon_1 = \varepsilon/3$ $\hspace{0.1in}$ and $\hspace{0.1in}$ $\varepsilon_2 = \varepsilon/3 \hspace{0.1in}$ then there exist

$\delta_1 > 0 \hspace{0.1in} $ such that $ \hspace{0.1in} \left \vert L \vert \cdot \vert {f\left( x \right) -K } \right| < \varepsilon_1 = \varepsilon/3 \hspace{0.2in}{\mbox{whenever}}\hspace{0.2in}0 < \left| {x - a} \right| < \delta_1 $

$\delta_2 > 0 \hspace{0.1in} $ such that $ \hspace{0.1in} \left \vert K \vert \cdot \vert {g\left( x \right) -L } \right| < \varepsilon_2 = \varepsilon/3 \hspace{0.2in}{\mbox{whenever}}\hspace{0.2in}0 < \left| {x - a} \right| < \delta_2 $

also choose $\varepsilon_3 = \sqrt{\varepsilon/3}$ $\hspace{0.1in}$ and $\hspace{0.1in}$ $\varepsilon_4 =\sqrt{\varepsilon/3} \hspace{0.1in}$ then there exist

$\delta_3 > 0 \hspace{0.1in} $ such that $ \hspace{0.1in} \left \vert {f\left( x \right) -K } \right| < \varepsilon_3 = \sqrt{\varepsilon/3} \hspace{0.2in}{\mbox{whenever}}\hspace{0.2in}0 < \left| {x - a} \right| < \delta_3 $

$\delta_4 > 0 \hspace{0.1in} $ such that $ \hspace{0.1in} \left \vert {g\left( x \right) -L } \right| < \varepsilon_4 = \sqrt{\varepsilon/3} \hspace{0.2in}{\mbox{whenever}}\hspace{0.2in}0 < \left| {x - a} \right| < \delta_4 $

choose the $\delta=\min(\delta_1,\delta_2,\delta_3,\delta_4)$ then :

$$ \begin{pmatrix} \hspace{0.1in} \left \vert L \vert \cdot \vert {f\left( x \right) -K } \right| < \varepsilon_1 = \varepsilon/3\\ \hspace{0.1in} \left \vert K \vert \cdot \vert {g\left( x \right) -L } \right| < \varepsilon_2 = \varepsilon/3\\ \hspace{0.1in} \left \vert {f\left( x \right) -K } \right| < \varepsilon_3 = \sqrt{\varepsilon/3}\\ \hspace{0.1in} \left \vert {g\left( x \right) -L } \right| < \varepsilon_4 = \sqrt{\varepsilon/3}\\ \end{pmatrix} $$ Will all hold for $\delta$

Adding the first two inequalities and Multiplying the last two inequalities we will get

$\hspace{.91in} { \vert L \vert \cdot \vert(f\left( x \right)-K) \vert +\vert K \vert \cdot \vert(g\left( x \right)-L)\vert{}+\vert f\left( x \right)-K \cdot g\left( x \right)-L\vert } < \varepsilon$ ${\mbox{whenever}}\hspace{2in}0 < \left| {x - a} \right| < \delta$

and that is what we looking for.

Ammar Bamhdi
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1 Answers1

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I think your proof is correct, but perhaps the choice of $\sqrt{\varepsilon/3}$ is not the most desirable, since you may need to show that if $\varepsilon$ is small, then $\sqrt\varepsilon$ is still small. In other words, that $\sqrt{x}\to 0$ when $x\to 0$.

To avoid such mess, the standard trick is to do as follows: $$|f(x)\cdot g(x) - LK| = |f(x)\cdot g(x) - f(x)\cdot L + f(x)\cdot L - KL| \leq $$ $$\leq |f(x)\cdot g(x) - f(x)\cdot L| + |f(x)\cdot L - KL|$$

Can you continue from here?

A. Salguero-Alarcón
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  • At first I think it is intuitively to consider $\sqrt\varepsilon$ will approach 0 as $\varepsilon$ approach 0 and by the restriction $\varepsilon$ > 0 we will guarantee also that any value under square root will be positive , for example look at the table :

    $$ \begin{array}{c|lcr} \varepsilon& \ \sqrt{\varepsilon/3} \ \hline 1 & 0.577 \ 0.1 & 0.182 \ 0.01 & 0.057 \ 0.001 & 0.0182\ 0.0001 & 0.00577\ ... & ... \end{array} $$

     – Ammar Bamhdi May 22 '19 at 21:25
  • and second, for the expression $ |f(x)\cdot g(x) - f(x)\cdot K| + |f(x)\cdot K - LK| = |f(x)|[(g(x)-K)]+|K|[f(x)-L]$

    Note that i have no control over$|g(x)-K|$ and $|f(x)-L|$

    thank You

     – Ammar Bamhdi May 22 '19 at 21:25
  • Yes, to the first comment, it is of course true, but perhaps it is not very common in this particular proof. To the second comment, I mistook K for L. I have corrected it now. – A. Salguero-Alarcón May 22 '19 at 21:28
  • $|f(x)\cdot g(x) - f(x)\cdot L| + |f(x)\cdot L - KL| = |f(x)| |(g(x)-L)| + |L| |f(x)-K|$

    OK but I do not know how to control $|f(x)|$ ?!

     – Ammar Bamhdi May 22 '19 at 21:35
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    $|f(x)| \leq |f(x) - K| + |K|$ – A. Salguero-Alarcón May 22 '19 at 21:51
  • Can you please verify form this :

    $|f(x)\cdot g(x) - LK| = |f(x)\cdot g(x) - f(x)\cdot L + f(x)\cdot L - KL|$ $ \hspace{1.4in}$$\leq |f(x)\cdot g(x) - f(x)\cdot L| + |f(x)\cdot L - KL|$

    $\hspace{1.4in}$$=|f(x)[g(x)-L]|+|L[f(x)-k]|$ $\hspace{1.4in}$$=|f(x)|.|g(x)-L|+|L|.|f(x)-k|$ $\hspace{1.4in}\leq|f(x)-k|+|k|.|g(x)-L|+|L|.|f(x)-k|$ $\hspace{1.4in}=|f(x)-K|[|K|.|g(x)-L|+|L|]$

    put $|f(x)-K| < \varepsilon/2$

    put $|K|.|g(x)-L|+|L| < 2$

    and by multiplying $\varepsilon/2\hspace{0.01in}.2 = \varepsilon$

     – Ammar Bamhdi May 23 '19 at 15:16
  • OK, you will not perhaps be able to guarantee that $|L| < 2$ (since $|L|$ is already given to you) but:
    • let $M>0$ be so that $|L|, |K| < M$
    • now let $|f(x)-K| < \frac{\varepsilon}{2M}$
     – A. Salguero-Alarcón May 23 '19 at 16:59
  • I think i done a mistake above, I Wrote the proof again:

    $|f(x)\cdot g(x) - LK| = |f(x)\cdot g(x) - f(x)\cdot L + f(x)\cdot L - KL|$ $ \hspace{1.4in}$$\leq |f(x)\cdot g(x) - f(x)\cdot L| + |f(x)\cdot L - KL|$

    $\hspace{1.4in}$$=|f(x)[g(x)-L]|+|L[f(x)-K]|$ $\hspace{1.4in}$$=|f(x)|.|g(x)-L|+|L|.|f(x)-K|$ $\hspace{1.4in}$$\leq|f(x)-K|+|K|.|g(x)-L|+|L|.|f(x)-K|$ $\hspace{1.4in}$$=|f(x)-K|(1+|K|.|g(x)-L|+|L|)$ $\hspace{1.4in}$$=|f(x)-K|(1+|K|.|g(x)|)$

    Put|M|>0 then $1+|K|.|g(x)|\leq |M|$

    Put $|f(x)-K|< |\varepsilon/M|$

    $\hspace{1.4in}$$=|f(x)-k|(1+|K|.|g(x)|)<|\varepsilon/M|.|M|$

     – Ammar Bamhdi May 23 '19 at 23:33
  • You made a mistake above. From $|f(x)|\cdot |g(x)-L| + |L| + |f(x)-K|$, let

    $|L|, |K| < M$ ;

    $|f(x)-L| < \frac{\varepsilon}{2M}$ ; and

    $|g(x)-M| < \frac{\varepsilon}{4M}$.
    Assume also that $\varepsilon < 1 < M$. Then $$|f(x)| \leq |f(x)-L| + |L| < \varepsilon + M < 2M$$ so $|f(x)|\cdot |g(x)-L| + |L| + |f(x)-K| \leq ...$

     – A. Salguero-Alarcón May 24 '19 at 08:41
  • OK I will Try to prove it with your new hints but can you tell me what exactly type of mistake that I've done with my proof to avoid it next time ? – Ammar Bamhdi May 24 '19 at 10:42
  • In your last step, you wrote $|g(x)-L| +|L| = |g(x)|$ – A. Salguero-Alarcón May 24 '19 at 15:48