# Thread: continuity of a function at a point

1. ## continuity of a function at a point

Which of the following functions f are continuous at the point a=3 and a=2.5:

f(x)=x if x is an integer.
f(x)=0 if x is not an integer.

I know the function is not continuous at a=3 and is continuous at a=2.5 but how do i prove this using Cauchy definition (epsilon-delta) of continuous functions.

Thank you.

2. Originally Posted by charikaar
Which of the following functions f are continuous at the point a=3 and a=2.5:

f(x)=x if x is an integer.
f(x)=0 if x is not an integer.

I know the function is not continuous at a=3 and is continuous at a=2.5 but how do i prove this using Cauchy definition (epsilon-delta) of continuous functions.

Thank you.
x=3:
If I give you a y some small distance (less than 1) away from 3, you get |f(x)-f(y)|=3. So if you pick epsilon to be 2.5, then can you find any interval that will get me $\displaystyle |f(x)-f(y)|<\epsilon$?

x=2.5:
Pick delta to be 0.4, what do you get?

3. Thank you for your help. I can now solve a=2.5 but still struggle with a=3 bit ..in our notes $\displaystyle |f(a+h)-f(a)|<\epsilon$ whats f(y) in your answer? thanks

4. Originally Posted by charikaar
Thank you for your help. I can now solve a=2.5 but still struggle with a=3 bit ..in our notes $\displaystyle |f(a+h)-f(a)|<\epsilon$ whats f(y) in your answer? thanks
The usual definition of continuity is; f is continuous at a point x if $\displaystyle \forall \epsilon >0, \exists \delta >0 \mbox{ s.t. } |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$. If you want to use your definition, think about what happens if $\displaystyle 0<h<1$. You don't need to worry about large distances away from a (if you prove it for small h, then if 0<h<"a number bigger than 1", then you pick h<1).