# Proving the multiplicative property of limits.

If $\lim_{x\rightarrow a}f(x) =L$ and $\lim_{x\rightarrow a}g(x) =M$ then prove $\lim_{x\rightarrow a}f(x)g(x) =LM$

Attempt

(Should there be any statement that I should write at start of this proof?) $$|f(x)g(x) -LM|=|f(x)g(x)+Lg(x)-Lg(x) -LM|=|g(x)[f(x)-L]+L[g(x)-M]<|g(x)|f(x)-L|+|L||g(x)-M|$$

Let $\epsilon>0$ then there exists $\delta_1, \delta_2$ such that $$|f(x)-L|<\frac{\epsilon}{2g(x)}~~ whenever~~ 0<|x-a|<\delta_1$$ $$|g(x)-M|<\frac{\epsilon}{2L}~~ whenever~~ 0<|x-a|<\delta_2$$

Choose $0<|x-a|<\delta$ where $\delta = \min(\delta_1,\delta_2)$ then we have $$|f(x)g(x) -LM|<\epsilon$$

Is this proof correct?

The reason I am concerned is in the second step where I defined the inequalities for the individual expression, I wrote it in terms of $g(x)$ and $L$. Is this step mathematically correct?

Edit:

Is this continuation correct?

Let $\epsilon>0$ then there exists $\delta_1, \delta_2$ such that $$|f(x)-L|<\frac{\epsilon}{2(1+|M|)}~~ whenever~~ 0<|x-a|<\delta_1$$ $$|g(x)-M|<\frac{\epsilon}{2(1+|L|)}~~ whenever~~ 0<|x-a|<\delta_2$$

Choose $0<|x-a|<\delta$ where $\delta = \min(\delta_1,\delta_2)$ then we have $$|f(x)g(x) -LM|<|g(x)|\frac{\epsilon}{2(1+|M|)}+|L|\frac{\epsilon}{2(1+|L|)}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

Fixed $0<\epsilon<1$. I would start from (your strict inequality in the first line of your proof is problematic) \begin{align} &|f(x)g(x) -LM|=|f(x)g(x)+Lg(x)-Lg(x) -LM|\\ \leqslant&\ |g(x)|\cdot|f(x)-L|+|L|\cdot|g(x)-M|\\ \leqslant&\ (|g(x)-M|+|M|)\cdot|f(x)-L|+|L|\cdot|g(x)-M| \end{align}

and show the following:

• there exists $\delta_1$ such that $$|f(x)-L|<\epsilon,\quad \hbox{for all }\ 0<|x-a|<\delta_1;$$

• there exists $\delta_2$ such that $$|g(x)-M|<\epsilon,\quad \hbox{for all }\ 0<|x-a|<\delta_2.$$

By choosing the smaller one as $\delta$, one has $$|f(x)g(x)-LM|\leqslant (1+|M|)\epsilon +|L|\epsilon \leqslant (1+|M|+|L|)\epsilon\tag{1}$$ for all $x$ with $0<|x-a|<\delta$.

[Added:] To see why (1) completes the proof, do the following exercise. Show that the following two statements are equivalent:

• There exists some constant $C$ such that for every $\epsilon>0$, $|A|\leqslant C\epsilon$.

• For every $\epsilon>0$, $|A|\leqslant \epsilon$.

Remark. One important tactic that is seldom mentioned in elementary real analysis textbooks is that when doing an estimate in analysis, one should never worry about the constant in front of one's epsilon. As Terry Tao points out in one of his excellent blog posts on problem solving strategies in real analysis:

Don’t worry too much about exactly what $\varepsilon$ (or $\delta$, or $N$, etc.) needs to be. It can usually be chosen or tweaked later if necessary.

• Is the statement proved? Arent we supposed to show $|f(x)g(x)-LM|<\epsilon$ – mathnoob123 Jul 15 '17 at 15:46
• Yes. This is the point of doing analysis. Let me explain later in my answer. – Jack Jul 15 '17 at 15:48
• Is this correct? $|A|<C\epsilon_1 = \epsilon_2$ – mathnoob123 Jul 15 '17 at 16:07
• I think you get the point. Need to be a little bit more careful about the writing though. – Jack Jul 15 '17 at 16:09
• So if I arise in such a situation such that I get $C\epsilon$ in the end. What statements should I write to show that this completes the proof? Is writing "$|A| = C\epsilon = \epsilon$ since $\epsilon$ is arbitrary" sufficient? – mathnoob123 Jul 15 '17 at 16:12

No, it is not correct. As you noticed, your $\delta_1$ also depends on $g(x)$. Besides, you divide by $g(x)$. What if $g(x)=0$?

• So what should be replaced over there? And what about the second expression? There I divided by 2L. What should be replaced there considering L could be 0? – mathnoob123 Jul 15 '17 at 15:09
• @FaiqRaees You should consider two cases: $M=0$ and $M\neq0$. Besides, you should find an upper bound $K$ of $g(x)$ near $a$. So, in order to prove that $|g(x)|.|f(x)-L|<\frac\epsilon2$, all you'll have to prove is that $K.|f(x)-L|<\frac\epsilon2$. – José Carlos Santos Jul 15 '17 at 15:13
• Please check the edit. – mathnoob123 Jul 15 '17 at 15:14
• @FaiqRaees I don't see where does the first $<$ sign from the last equality come from. – José Carlos Santos Jul 15 '17 at 15:18
• Can you please be a little more specific? – mathnoob123 Jul 15 '17 at 15:20

Your approach is fine but a few details are not correct.

You should replace $$|f(x)-L|<\frac{\epsilon}{2g(x)}$$ with something like $$|f(x)-L|<\frac{\epsilon}{2M'}.$$ where $M'$ is a positive constant to be decided. Recall that $\lim_{x\rightarrow a}g(x) =M$ implies that $g$ is bounded in a neighbourhood of $a$.

Moreover to avoid problems with the case when $L=0$, replace $$|g(x)-M|<\frac{\epsilon}{2L}$$ for example with $$|g(x)-M|<\frac{\epsilon}{2(1+|L|)}.$$

Are you able to finish to proof correctly now?

P.S. In your edited work add th following line $|g(x)|<{(1+|M|)}$ whenever $0<|x-a|<\delta_3$ and define $\delta = \min(\delta_1,\delta_2,\delta_3)$.

• Okay what about the second expression? There I divided by 2L. Would should be replaced there considering L could be 0? – mathnoob123 Jul 15 '17 at 15:06
• @Faiq Raees See my edited answer. Are you able to finish to proof correctly now? – Robert Z Jul 15 '17 at 15:13
• Yes. Please check the edit – mathnoob123 Jul 15 '17 at 15:14
• @Faiq Raees You are almost done. Add one line $|g(x)|<{(1+|M|)}$ whenever $0<|x-a|<\delta_3$ and $\delta = \min(\delta_1,\delta_2,\delta_3)$ – Robert Z Jul 15 '17 at 15:16
• Actually, I have some concerns over this substitution. We are working on the greater side of inequality. If we substitute something that is greater than gx as denominator, it will make the fraction smaller, and the ineqaulity might not hold anymore. – mathnoob123 Jul 15 '17 at 15:25