# tough continuity problem need help

• Dec 6th 2006, 09:39 AM
myoplex11
tough continuity problem need help
g(x)=0 if x is rational
g(x)=x if x is irrational

for what values of x is g continuous?
• Dec 6th 2006, 09:44 AM
ThePerfectHacker
Quote:

Originally Posted by myoplex11
g(x)=0 if x is rational
g(x)=x if x is irrational

for what values of x is g continuous?

Have you ever heard of "Dirichelt" function?
Same idea.

The limit that we approach a point bt rationals needs to be the same as we approach the limit by irrationals. In that case $\displaystyle x=0$ is the only continous point.
Look Heir
• Dec 6th 2006, 10:30 PM
myoplex11
i understand why it is continous a x=0 but i need to justify this mathematically
• Dec 7th 2006, 04:37 AM
CaptainBlack
Quote:

Originally Posted by myoplex11
i understand why it is continous a x=0 but i need to justify this mathematically

Choose any $\displaystyle \delta>0$ and set $\displaystyle \epsilon=\delta$, then $\displaystyle |g(x)-g(0)|<\delta$ whenever $\displaystyle |x|<\epsilon$,
hence $\displaystyle g(x)$ is continuouse at $\displaystyle x=0$.

RonL