Proof of the Product Limit Law

Question

Theorem:

$$\lim_{x \to a} f(x) = L$$

$$\lim_{x \to a} g(x) = M$$

Then:

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

Obviously,

$$|f(x) - L| < \epsilon$$

$$|g(x) - M| < \epsilon$$

But multiplying these together doesnt get the desired:

$$|f(x)g(x) - LM| < \epsilon$$

Please, HINTS only!

Answer 1

Hint:

$$\begin{align}|f(x)g(x)-LM|&=|f(x)g(x)-Lg(x)+Lg(x)-LM|\\&=|g(x)(f(x)-L)+L(g(x)-M)|\\&\le|g(x)||f(x)-L|+|L||g(x)-M|\end{align}$$

Answer 2

Prove it for the special case where the two functions are infinitesimal as $x\to a$ that is $\lim_{x\to a}f(x)=0$

Prove the theorem for such special functions then deduce the theorem.

Answer 3

Theorem 1:

Denote $f(x)=L+\alpha(x)$.

$$\lim_{x\to a}f(x)=L\iff \lim_{x\to a}\alpha(x)=0$$

Proof 1:

Denote $f(x)=L+\alpha(x)$.

$$\lim_{x\to a}f(x)=L\iff \forall\epsilon >0(\exists\delta>0(0<|x-a|<\delta\implies |f(x)-L|<\epsilon)$$

$$\iff \forall\epsilon >0(\exists\delta>0(0<|x-a|<\delta\implies |\alpha(x)|<\epsilon)\iff \lim_{x\to a}\alpha(x)=0\ \ \ \square$$

$$f(x)g(x)=(L+\alpha(x))(M+\beta(x))=LM+\beta(x)L+M\alpha(x)+\alpha(x)\beta(x)$$

$\lim_{x\to a}f(x)g(x)=LM$ holds iff $$\lim_{x\to a}(\beta(x)L+M\alpha(x)+\alpha(x)\beta(x))=0$$

Theorem 2:

$$\lim_{x\to a}\alpha(x)B(x)=0,$$

where $\lim_{x\to a}\alpha(x)=0$ and $\exists m,M\in\mathbb R, m,M\neq 0(m\le B(x)\le M)$ (so that $|B|\le \max\{|m|,|M|\}$)

Proof 2:

$$\lim_{x\to a}\alpha(x)B(x)=0\iff\forall\epsilon>0(\exists\delta>0(0<|x-a|<\delta\implies |\alpha(x)B(x)|<\epsilon))$$

We know that $$\lim_{x\to a}\alpha(x)=0\iff \forall\epsilon>0(\exists\delta>0(0<|x-a|<\delta\implies |\alpha(x)|<\epsilon))$$

$$|\alpha(x)B(x)|=|\alpha(x)||B(x)|\le |\alpha(x)|\cdot\max\{|m|,|M|\}<\epsilon\cdot\max\{|m|,|M|\}$$

$\epsilon\cdot\max\{|m|,|M|\}$ can obtain any positive real value, because any real positive value $k$ is obtained when $\epsilon=\frac{k}{\max\{|m|,|M|\}}>0$. $\ \ \ \square$

Theorem 3:

$$\lim_{x\to a}(\alpha(x)+\beta(x))=0,$$

where $\lim_{x\to a}\alpha(x)=0, \lim_{x\to a}\beta(x)=0$.

Proof 3:

$$\lim_{x\to a}\alpha(x)=0\iff\forall\epsilon_1>0(\exists\delta>0(0<|x-a|<\delta\implies |\alpha(x)|<\epsilon_1))$$

$$\lim_{x\to a}\beta(x)=0\iff\forall\epsilon_2>0(\exists\delta>0(0<|x-a|<\delta\implies |\beta(x)|<\epsilon_2))$$

Since they both hold for any $\epsilon_1, \epsilon_2>0$, take $\frac{\epsilon}{2}=\epsilon_1=\epsilon_2$.

Then

$$|\alpha(x)+\beta(x)|\le |\alpha(x)|+|\beta(x)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

So that

$$\forall\epsilon>0(\exists\delta>0(0<|x-a|<\delta\implies |\alpha(x)+\beta(x)|<\epsilon))\ \ \ \square$$

Our statement

$$\lim_{x\to a}(\beta(x)L+M\alpha(x)+\alpha(x)\beta(x))=0$$

is trivial using Theorem 2 and Theorem 3. $\ \ \ \square$