# Prove that if $\lim_{x\rightarrow c}f(x)=L$ and $\lim_{x\rightarrow c}g(x)=M$ then $\lim_{x\rightarrow c}f(x)g(x)=LM$.

We have $$\lim_{x\rightarrow c}f(x)=L \\ \lim_{x\rightarrow c}g(x)=M$$ We need to show the following using the $\epsilon-\delta$ definition of limits. $$\lim_{x\rightarrow c}f(x)g(x)=LM$$

By the definition we have for $|x-c|<\delta_1, |x-c| < \delta_2$ $$L-\epsilon_1<f(x)<L+\epsilon_1 \\ M-\epsilon_2<g(x)<M+\epsilon_2$$ Multiplying the two $$LM-L\epsilon_2-M\epsilon_1+\epsilon_1\epsilon_2 < f(x)g(x)<LM+L\epsilon_2+M\epsilon_1+\epsilon_1\epsilon_2$$ We ignore $\epsilon_1\epsilon_2$ as it can be very small for small enough $\epsilon$. And we get $$|f(x)g(x)-LM|<L\epsilon_2+M\epsilon_1=\epsilon$$ Which holds for $|x-c|<\min(\delta_1, \delta_2)$.

I want to know if it is okay to drop the $\epsilon_1\epsilon_2$. Is there a better way to prove this?

• You cannot multiply inequalities. For example, $-10<-1,-10<-2$ but $(-10)(-10) >(-1)(-2)$. Apr 4 '18 at 5:33

As I mentioned in my comment your argument fails completely. Take $\epsilon _1=\epsilon_2 =\delta$ where $0<\delta<\epsilon/(|L|+|M|+1)$ and use the inequalities $|f(x)g(x)-LM| \leq |f(x)g(x)-f(x)M|+|f(x)M-LM|<(|L|+\delta)\delta+\delta |M| <\epsilon$ for $|x-c|< \min \{\delta_1,\delta_2\}$.