# Thread: Continuous

1. ## Continuous

Let f be a function such that f(x)=x^2 for all numbers x. Show that f is continuous at the point (2,4).

2. Originally Posted by spikedpunch
Let f be a function such that f(x)=x^2 for all numbers x. Show that f is continuous at the point (2,4).
Notice that $|x^2-4|=|x-2||x+2|$.
If $|x-2|<1$ then $|x+2|<5$.
$\varepsilon > 0\, \Rightarrow \,\delta = \min \left\{ {1,\varepsilon /5} \right\}$

3. could you explain why delta is min( 1, epsilon/5 )?

4. Originally Posted by spikedpunch
could you explain why delta is min( 1, epsilon/5 )?
Well you want it to work.
$\left| {x - 2} \right| < \delta = \min \left\{ {1,\varepsilon /5} \right\}\, \Rightarrow \,\left| {x + 2} \right| < 5$

$\left| {x^2 - 4} \right| = \left| {x + 2} \right|\left| {x - 2} \right| < 5\left( {\varepsilon /5} \right) = \varepsilon$

5. So I can see this works if if the min is epsilon/5, but how would it work if the min is 1? That is, how can I show that the f(x)-f(t)<epsilon?

6. Originally Posted by spikedpunch
[COLOR=#0000AA][FONT=monospace] $
\left| {x^2 - 4} \right| = \left| {x + 2} \right|\left| {x - 2} \right| < 5\left( {\varepsilon /5} \right) = \varepsilon$

So I can see this works if if the min is epsilon/5, but how would it work if the min is 1?
Just think about it?
If $t=\min\{1,2\}$ then $t\le1~\&~t\le 2$, right?

7. I guess I am confused as to why you showed -1<x-2<1 in the first place. If 1 is the min, then wouldn't it be

absv(x^2-4)=absv(x+2)absv(x-2)=(1)(5)=5...which is not epsilon?

8. Originally Posted by spikedpunch
I guess I am confused as to why you showed -1<x-2<1 in the first place. If 1 is the min, then wouldn't it be absv(x^2-4)=absv(x+2)absv(x-2)=(1)(5)=5...which is not epsilon?
I think that you need to do some background work on this.