$\epsilon$-$\delta$ proof of $\lim f(x)g(x) = \lim f(x)\lim g(x)$ for $x\to a$

Question

Can someone please hold my hand and guide me through this proof. I saw it in spivaks examples and yet it does not make sense! Your help is very appreciated

Answer 1

Denote $$\lim_{x\to a}f(x)=\ell\quad\text{and}\quad \lim_{x\to a}g(x)=\ell'$$

and by the limit's definition we have for $\epsilon>0$ there's $\delta,\delta'>0$ $$|f(x)-\ell|<\epsilon\quad \text{if}\; |x-a|<\delta$$ and $$|g(x)-\ell'|<\epsilon\quad \text{if}\; |x-a|<\delta'$$ and for $\delta''=\min(\delta,\delta')$ we have $$|f(x)g(x)-\ell\ell'|=|f(x)g(x)-g(x)\ell+g(x)\ell-\ell\ell'|\le|g(x)|\cdot|f(x)-\ell|+|\ell|\cdot|g(x)-\ell'|\\\le (M+|\ell|)\epsilon \quad \text{if}\; |x-a|<\delta''$$ where $|g(x)|<M$ for $|x-a|<\delta''$ since $g$ has a limit at $a$ then it's bounded in a neighborhood of $a$. It remains for us to conclude.

Answer 2

Assuming that $f$ and $g$ are continuous at the point $x=a$, we have that

$|f(x)g(x) - f(a)g(a)| \leq |f(x)g(x) - f(a)g(x) + f(a)g(x) - f(a)g(a)|$ $\leq |g(x)||f(x) - f(a)| + |f(x)||g(x) - g(a)|$

Now since $f,g$ are continuous at $x=a$, the function $f,g$ are bounded in a neighborhood about $x=a$, call this bound $M$. Moreover, given $\epsilon > 0$ there exists $\delta > 0$ such that $|f(x) - f(a)|, |g(x) - g(a)| \leq \frac{\epsilon}{2M}$ when $|x-a| < \delta$. Hence, using the inequalities above, we get that $|f(x)g(x) - f(a)g(a)| \leq 2M(\frac{\epsilon}{2M}) \leq \epsilon$ as when $|x-a| < \delta$, as needed.

Answer 3

Assume that $a$ is a limit point of the domain of $f$ and $g$ which is $A\subseteq X$.

\begin{align} \lim_{x\to a}f(x) &= f\\ \therefore\mbox{Given }\epsilon>0 & \exists\delta_{f}>0\\ 0<|x-a|<\delta_{f} &\implies |f(x)-f|<\frac{\epsilon}{2M} \qquad\mbox{Explanation Below}\\ \lim_{x\to a}g(x) &= g\\ 0<|x-a|<\delta_{g} &\implies |g(x)-g|<\frac{\epsilon}{2(|f|+1)}\\ \mbox{Now, } \forall x\;s.t. &0<|x-a|<\min\{\delta_f,\delta_g\}\\ |f(x)g(x)-fg| &= |f(x)g(x)-fg(x)+fg(x)-fg|\\ &\leq |f(x)g(x)-fg(x)|+|fg(x)-fg|\\ &= |f(x)-f|\cdot|g(x)|+|f|\cdot|g(x)-g|\\ &< \frac{\epsilon}{2M}\cdot|g(x)|+|f|\cdot\frac{\epsilon}{2(|f|+1)}\\ &< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\\ \end{align}

Explanation for M : Because $\lim_{x\to a}g(x)$ exists, If $(x_{n})$ is a sequence converging to $a$ in $A\setminus a$ , $g(x_{n})$ converges to $g$ . Convergent sequences are bounded. Hence, $|g(x)|<M$ .