0
$\begingroup$

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?

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

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 $|x-c|< \min \{\delta_1,\delta_2\}$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.