# Limit Law Questions

• Sep 16th 2009, 08:30 PM
amma0913
Limit Law Questions
Hi everyone, I need some help with two limit law questions.

1. Is it possible for the limit of f(x)*g(x) as x-->a to exist even if lim f(x) as x-->a and lim g(x) x-->a do not exist?

2. Is it possible for the limit of f(x)+g(x) as x-->a to exist even if lim f(x) as x-->a and lim g(x) as x-->a do not exist?

I know the first one is true, but I can't think of an example. Could someone please explain if either question is true or false and give an example that justifies their reason?

Thanks.
• Sep 17th 2009, 05:03 AM
HallsofIvy
Quote:

Originally Posted by amma0913
Hi everyone, I need some help with two limit law questions.

1. Is it possible for the limit of f(x)*g(x) as x-->a to exist even if lim f(x) as x-->a and lim g(x) x-->a do not exist?

I cannot think of an example where both limits do not exist by limit of f(x)g(x) does but if f(x)= x, g(x)= 1/x, the limit of g(x), as x goes to 0, does not exist while f(x)g(x)= x(1/x)= 1 so its limit is 1.

Quote:

2. Is it possible for the limit of f(x)+g(x) as x-->a to exist even if lim f(x) as x-->a and lim g(x) as x-->a do not exist?
Look at f(x)= 1/x, g(x)= -1/x, as x goes to 0.

Quote:

I know the first one is true, but I can't think of an example. Could someone please explain if either question is true or false and give an example that justifies their reason?

Thanks.
• Sep 17th 2009, 06:19 AM
CaptainBlack
Quote:

Originally Posted by amma0913
Hi everyone, I need some help with two limit law questions.

1. Is it possible for the limit of f(x)*g(x) as x-->a to exist even if lim f(x) as x-->a and lim g(x) x-->a do not exist?

2. Is it possible for the limit of f(x)+g(x) as x-->a to exist even if lim f(x) as x-->a and lim g(x) as x-->a do not exist?

I know the first one is true, but I can't think of an example. Could someone please explain if either question is true or false and give an example that justifies their reason?

Thanks.

Let:

$
f(x)= \begin{cases} 0 & x \text{ rational } \\ 1 & x \text{ irrational }
\end{cases}
$

and:

$
g(x)= \begin{cases} 1 & x \text{ rational } \\ 0 & x \text{ irrational }
\end{cases}
$

CB