# Limit of Product Proof

I have to show that if $\lim_{x \to a}f(x) = L$ and $\lim_{x \to a}g(x) = M$, $\lim_{x \to a} f(x)g(x) = LM$. I came up with a proof but it's different from the ones I've found in my textbook or on the internet. It seems right to me, but I'm still new to this.

We want to show that there exists a $\delta$ such that if $|x - a| < \delta$, $|f(x)g(x) - LM| < \varepsilon$. By the definition of a limit, there exists a $\delta_1$ such that if $|x - a| < \delta_1$, $|f(x) - L| < \min(1, \frac{\varepsilon}{2|M|}) = \varepsilon_1$. Similarly, there exists a $\delta_2$ such that if $|x - a| < \delta_2$, $|g(x) - M| < \frac{\varepsilon}{2(|L| + 1)} = \varepsilon_2$.

Let $\delta = \min(\delta_1, \delta_2)$.

This means $$|Mf(x) - LM| < \frac{\varepsilon}{2}$$ $$|Lg(x) - LM| < \frac{|L|\varepsilon}{2(|L| + 1)}$$ and $$|f(x)g(x) - Lg(x) - Mf(x) + LM| = |f(x) - L||g(x) - M| < \varepsilon_1\varepsilon_2 < \varepsilon_2$$ Combining these yields $$|f(x)g(x) - Lg(x) - Mf(x) + LM| + |Mf(x) - LM| + |Lg(x) - LM| < \frac{\varepsilon}{2} + \frac{|L|\varepsilon}{2(|L| + 1)} + \frac{\varepsilon}{2(|L| + 1)}$$ Which simplifies to $$|f(x)g(x) - LM| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

• If $M=0$, then $\epsilon_1$ isn't defined. If $L=0$, then the second inequality after the definition of $\delta$ isn't true. – JDZ May 11 '17 at 5:08
• $\varepsilon_1$ can be defined as $\min(1, \frac{\varepsilon}{2(|M|+1)})$ to include the case $M=0$. Also, the inequality may be rendered as $|Lg(x) - LM| \le \frac{|L|\varepsilon}{2(|L| + 1)}$ which the equality holds when $L=0$. I think the proof shall work then. – delt31 May 11 '17 at 5:31

Put $h(x) = f(x) - L, k(x) = g(x) - M$, then the problem becomes $h(x) \to 0, k(x) \to 0$ as $x \to a$, and the statement $fg \to LM$ becomes $(h+L)(k+M) \to LM$ or $hk + hM + kL \to 0$. The case $LM = 0$ should be considered first and it is easy to deal with it. So if $LM \neq 0$, then you can make $|hk| < \dfrac{\epsilon}{3}, |h| < \dfrac{\epsilon}{3|M|}, |k| < \dfrac{\epsilon}{3|L|}$, and using triangle inequality to get $|hk+hM+kL| < \epsilon$. Can you work out the details?
Given $\epsilon>0,$ let $\epsilon'=\min \left(1,\frac {\epsilon}{|M|+|L|+1}\right).$
Let $f(x)=L+f^*(x)$ and $g(x)=M+g^*(x).$
Take $\delta >0$ such that $|x-a|<\delta\implies [\;(|f(x)-L|<\epsilon')\;\land \;(|g(x)-M|<\epsilon')\;]$. That is, $$|x-a|<\delta \implies [\;(|f^*(x)|<\epsilon')\;\land \;(|g^*(x)|<\epsilon')\;\land \;(|f^*(x)|<1)\;].$$ Now we have $$|x-a|<\delta \implies |f(x)g(x)-LM|=$$ $$=|(L+f^*(x))(M+g^*(x))-LM|=$$ $$= |Mf^*(x)\;+\;Lg^*(x)\;+\;f^*(x)g^*(x)|\leq$$ $$\leq |M|\cdot |f^*(x)|\;+\;|L|\cdot |g^*(x)|\;+\;|f^*(x)|\cdot |g^*(x)|\leq$$ $$\leq |M|\epsilon'\;+\;|L|\epsilon'\;+\;1\cdot |g^*(x)| <$$ $$< (|M|+|L|+1)\epsilon'\;\leq\; \epsilon .$$
Note: We can find $\delta_1>0$ such that $|x-a|<\delta_1\implies |f(x)-L|<\epsilon'/2.$ And we can find $\delta_2>0$ such that $|x-a|<\delta_2\implies |g(x)-M|<\epsilon'.$ So let $\delta=\min (\delta_1, \delta_2).$