# function continuous at a point

• Apr 12th 2010, 06:47 AM
charikaar
function continuous at a point
f:R-->R
f(x)=1 if $\displaystyle x = 0$
f(x)=0 otherwise.
(a) Is f continuous at the point a = 0?
(b) Is f continuous at the point a = 1?

have i done them correctly?

Continuous at a=0 because $\displaystyle \lim_{x\to0}f(x)=1=f(0)$ correct?

also continuous at a=1 because $\displaystyle \lim_{x\to1}f(x)=0=f(1)$

thanks
• Apr 12th 2010, 11:09 AM
Focus
Quote:

Originally Posted by charikaar
f:R-->R
f(x)=1 if $\displaystyle x = 0$
f(x)=0 otherwise.
(a) Is f continuous at the point a = 0?
(b) Is f continuous at the point a = 1?

have i done them correctly?

Continuous at a=0 because $\displaystyle \lim_{x\to0}f(x)=1=f(0)$ correct?

also continuous at a=1 because $\displaystyle \lim_{x\to1}f(x)=0=f(1)$

thanks

Are you really sure of the first case? When you are not at 0 (which you aren't when you are taking a limit), the function takes the value 0. What does $\displaystyle x_n=0$ converge to?

I mean you can think about it more intuitively, is the function "connected" at 0? Draw it.
• Apr 12th 2010, 11:48 AM
charikaar
I see what you mean. The function is not connected at zero and hence the limit in first case doesn't exist so not continuous at a=0
• Apr 12th 2010, 01:44 PM
Focus
Quote:

Originally Posted by charikaar
I see what you mean. The function is not connected at zero and hence the limit in first case doesn't exist so not continuous at a=0

The limit does exist! Try to compute it this way
$\displaystyle f(1/2), f(1/4), f(1/8),...$
what it the limit of this sequence? (it's the same as $\displaystyle \lim_{x \downarrow 0} f(x)$).

I think it may help if you look up the definition of a limit. The limit at a=0 might not be f(a) (which is the case here), but indeed it still does exist.