# Math Help - continuity question

1. ## continuity question

Regarding
$
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?]

2. I think youve made a mistake in that, I think it should say $f(x) = x^2$ 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.

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

There's some pictures of these functions on wikipedia

3. 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]?"

4. 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

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

What are those two limits?

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 $\lim_{x\to x_0} 0= 0$?

6. Sorry but I just realized I wrote the function wrong, here is the correct form:

$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 $x_{0}\in [1,2]$, then $\forall \epsilon >0, \exists \delta >0$ such that $|f(x)-f(x_{0})|<\epsilon$ provided that $|x-x_{0}|<\delta$ and $x\in[1,2]$
Actually Im not sure if x must exist in [1,2], any comment?

7. So I let $\delta=\frac{\epsilon}{2}$ 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?