Verifying form the proof of product theorem of limit

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.

1 Answer

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?

• 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 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 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 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 at 21:35
• $|f(x)| \leq |f(x) - K| + |K|$ – A. Salguero-Alarcón May 22 at 21:51