# continuity question

• Oct 22nd 2009, 10:24 PM
binkypoo
continuity question
Regarding
$\displaystyle f(x) = \left\{ \begin{array}{lr} x^2 & : x < 0\\ 0 & : 0 \leq x \leq 2 \end{array} \right.$
Is f continuous on [0,1] or [1,2]?
and is f 'right continuous' on [0,1]?
and why?

[I dont really know what it means for a function to be 'right continuous'...
any help on this?]
• Oct 23rd 2009, 01:18 AM
slevvio
I think youve made a mistake in that, I think it should say $\displaystyle f(x) = x^2$ for x < 0.

If a function is right continuous then the limit of f(x) as you approach a from the right is f(a).

i.e.

$\displaystyle \forall \epsilon > 0, \exists \delta > 0$ such that $\displaystyle \forall x \in (a, a+\delta), |f(x) - f(a) | < \epsilon$

There's some pictures of these functions on wikipedia
• Oct 23rd 2009, 04:54 AM
HallsofIvy
Note that for a function to be "continuous" at a point, it must be both "right continuous" and "left continuous".

What was your answer to the question "Is f continuous on [0,1] or [1,2]?"
• Oct 23rd 2009, 11:36 AM
binkypoo
I believe f is not continuous on [0,1] because the right side limit and left side limit at 0 are not equal.
on [1,2] it seems to me that it is continuous but Im not sure how to prove either of these claims
• Oct 23rd 2009, 06:00 PM
HallsofIvy
Quote:

Originally Posted by binkypoo
I believe f is not continuous on [0,1] because the right side limit and left side limit at 0 are not equal.

The "right side limit" is $\displaystyle \lim_{x\to 0^+} f(x)= \lim_{x\to 0} 0$ and the "left side limit" is $\displaystyle \lim_{x\to 0^-} f(x)= \lim_{x\to 0} x^2$.

What are those two limits?

Quote:

on [1,2] it seems to me that it is continuous but Im not sure how to prove either of these claims
on [1, 2], f(x) is identically 0. Can't you prove that $\displaystyle \lim_{x\to x_0} 0= 0$?
• Oct 25th 2009, 07:58 PM
binkypoo
Sorry but I just realized I wrote the function wrong, here is the correct form:

$\displaystyle f(x) = \left\{ \begin{array}{lr} x^2 & : x < 0\\ 2x + 1 & : 0 \leq x \leq 2 \end{array} \right.$
Im stumped on proving continuity on the interval [1,2]. I can prove continuity at a point but dont know how to do it on a whole interval

Heres what I have so far:
I want to show if $\displaystyle x_{0}\in [1,2]$, then $\displaystyle \forall \epsilon >0, \exists \delta >0$ such that $\displaystyle |f(x)-f(x_{0})|<\epsilon$ provided that $\displaystyle |x-x_{0}|<\delta$ and $\displaystyle x\in[1,2]$
Actually Im not sure if x must exist in [1,2], any comment?
• Oct 25th 2009, 10:56 PM
binkypoo
So I let $\displaystyle \delta=\frac{\epsilon}{2}$ to show f is continuous on [1,2] But I feel like delta should somehow involve the interval [1,2]... not sure where to go. any suggestions?