# Continuity

• Aug 2nd 2010, 05:46 PM
novice
Continuity
For function $\displaystyle f$ to be continuous at $\displaystyle a$, three conditions must be satisfied:

(1)$\displaystyle f$ is defined at $\displaystyle a$;
(2) $\displaystyle \lim_{x \to a} f(x)$ exists;
(3) $\displaystyle \lim_{x \to a} f(x)=f(a)$ exists;

Now say a function $\displaystyle g$ is defined as

$\displaystyle g(x) = \left \{\begin{array}{cc}0,&\mbox{if} x\not \in \mathbb{Z}\\1,&\mbox{if} x\in \mathbb{Z} \end{array}\right$.

Can we say $\displaystyle g$ is continuous $\displaystyle a$ where $\displaystyle a=2$ and $\displaystyle a \in \mathbb{Z}$?

Does $\displaystyle g$ satisfy the following?

(1)$\displaystyle f$ is defined at $\displaystyle a$ where $\displaystyle f(2)=1$;
(2) $\displaystyle \lim_{x \to a} f(x)$ exists;
(3) $\displaystyle \lim_{x \to a} f(x)=f(a)=0$ exists

?
• Aug 2nd 2010, 06:31 PM
eumyang
Quote:

Originally Posted by novice
Now say a function $\displaystyle g$ is defined as

$\displaystyle g(x) = \left \{\begin{array}{cc}0,&\mbox{if} x\not \in \mathbb{Z}\\1,&\mbox{if} x\in \mathbb{Z} \end{array}\right$.

Can we say $\displaystyle g$ is continuous at $\displaystyle a$ where $\displaystyle a=2$ and $\displaystyle a \in \mathbb{Z}$?

Does $\displaystyle g$ satisfies the following?

(1)$\displaystyle g(a)$ is defined.
Yes, $\displaystyle g(2)=1$.

(2) $\displaystyle \lim_{x \to a} g(x)$ exists;
Yes, $\displaystyle \lim_{x \to 2} g(x) = 0$.

(3) $\displaystyle \lim_{x \to a} g(x)=g(a)$
No. We have in (1) that $\displaystyle g(2)=1$ and in (2) $\displaystyle \lim_{x \to 2} g(x) = 0$. They're not equal, so condition (3) isn't satisfied. So g is not continuous at x = 2.
• Aug 3rd 2010, 06:44 AM
novice
Now, change point $\displaystyle a$ to $\displaystyle a \not \in \mathbb{Z}$.

Then $\displaystyle g(x)$ is contiunous at $\displaystyle x= a, \forall a \not \in \mathbb{Z}$.

Yah?
• Aug 3rd 2010, 06:48 AM
Plato
Yes indeed.