Proving continuity at a point.

**chr91** · December 3rd 2010, 02:29 AM

Show that f(x) = x^3 is continuous at 1 (from the definition)

(For all epsilon >0) (There exists delta >0) (For all x E X) if |x-1| < delta this implies
|x^3 - 1^3| < epsilon

|x^3 - 1^3| = |x-1| |x^2 + x + 1|

I've not got any further with this. Could I grab some help!

**hmmmm** · December 3rd 2010, 02:31 AM

so you have factored out x-a so now you want to use delta to put a bound on that, you then know what x will vary between and so you will know the max value of |x^2+x+1|, does that help?

$\displaystyle\mid(f(x)-L)$

$\displaystyle\mid(x^3-1)\mid=\mid(x-1)\mid\mid(x^2+x+1)\mid$

We then use $\displaystyle\delta$ to bound $\displaystyle\mid(x-1)\mid$

so let $\displaystyle\delta=1\rightarrow\mid(x-1)\mid\le1$ so we know that x varies between 0 and 2

so $\displaystyle\((x^2+x+1)\le7$

**chr91** · December 3rd 2010, 02:34 AM

Originally Posted by hmmmm

so you have factored out x-a so now you want to use delta to put a bound on that, you then know what x will vary between and so you will know the max value of |x^2+x+1|, does that help?

Erm, not really. I know I need to find the delta but I'm not sure what ..

**DrSteve** · December 3rd 2010, 02:45 AM

If a given delta works, then any smaller delta works as well. Thus, you may assume that $|x - 1|$ is less than some fixed value, let's say $|x-1|<1$

Then $-1 < x-1 < 1$ , and so $0 < x < 2$ .

So, $x^2 + x + 1 < 2^2 + 2 + 1 = 7$ , and $x^2 + x + 1 > 0^2 + 0 + 1 = 1$ . So $-7 < 1 < x^2 + x + 1 < 7$ , or equivalently $|x^2 + x + 1| < 7$

**chr91** · December 3rd 2010, 02:47 AM

Originally Posted by hmmmm

so you have factored out x-a so now you want to use delta to put a bound on that, you then know what x will vary between and so you will know the max value of |x^2+x+1|, does that help?

$\displaystyle\mid(f(x)-L)$
$\displaystyle\mid\=\mid(x^3-1)\mid\=\mid(x-1)\mid\mid(x^2+x+1)\mid$

We then use $\displaystyle\delta$ to bound $\displaystyle\mid(x-1)\mid$
so let $\displaystyle\delta=1\rightarrow\mid(x-1)\mid\le1$ so we know that x varies between -1 and 2
so $\displaystyle\(x^2+x+1)\le7$

If delta = 1, and |x-1| < 1 then surely x varies between 0 and 2?

**chr91** · December 3rd 2010, 02:52 AM

Ok so this is my proof:

Let |x-1| < 1 so 0<x<2

Therefore |x^2 + x + 1| < 7

Let delta = epsilon/7

Suppose |x-1| < delta
|x^3 - 1^3| = |x-1||x^2 + x + 1| < 7|x-1| < 7delta = epsilon

Correct?

**DrSteve** · December 3rd 2010, 03:05 AM

Almost - you have to let delta = min{1, epsilon/7}.

This is because in your argument you assumed that delta was less than 1. So if epsilon/7 turns out to be bigger than 1, then delta = epsilon/7 doesn't work.

**chr91** · December 3rd 2010, 03:07 AM

Originally Posted by DrSteve

Almost - you have to let delta = min{1, epsilon/7}.

This is because in your argument you assumed that delta was less than 1. So if epsilon/7 turns out to be bigger than 1, then delta = epsilon/7 doesn't work.

Oh yeh, must be careful with that.

Thanks!

**hmmmm** · December 3rd 2010, 03:15 AM

yeah sorry i just made a little mistake there but i was of no consequence,
let $\displaystyle\epsilon$ be given choose $\displaystyle\delta=min\{1,\epsilon/7\}$

We then know that if $\displaystyle{0}\le\mid(x-1)\mid\le\delta\le{1}$
then
$\displaystyle\mid(x^3-1)\mid\le7\mid(x-1)\mid$

so we conclude that

$\displaystyle0\le\mid(x-1)\mid\le\delta\rightarrow\mid(x^3-1)\mid\le7*\mid(x-1)\mid\le7*\frac{\epsilon}{7}=\epsilon$

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