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 $\displaystyle |x^2-4|=|x-2||x+2|$.
If $\displaystyle |x-2|<1$ then $\displaystyle |x+2|<5$.
$\displaystyle \varepsilon > 0\, \Rightarrow \,\delta = \min \left\{ {1,\varepsilon /5} \right\}$

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.
$\displaystyle \left| {x - 2} \right| < \delta = \min \left\{ {1,\varepsilon /5} \right\}\, \Rightarrow \,\left| {x + 2} \right| < 5$

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

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]$\displaystyle \left| {x^2 - 4} \right| = \left| {x + 2} \right|\left| {x - 2} \right| < 5\left( {\varepsilon /5} \right) = \varepsilon$
So I can see this works if if the min is epsilon/5, but how would it work if the min is 1?
If $\displaystyle t=\min\{1,2\}$ then $\displaystyle t\le1~\&~t\le 2$, right?