# Having trouble showing continuity.

• Oct 30th 2012, 11:57 AM
pantsaregood
Having trouble showing continuity.
I wish to show f(x)=x3 such that x is an element of the real numbers is continuous at an arbitrary x0.

This means that, given some epsilon greater than zero (henceforth referred to simply as "E"), there exists a delta greater than zero (henceforth referred to as "D") such that...

|x-x0|<D implies |f(x)-f(x0)|<E

So, we have...

|x-x0|<D should imply |x3-x03|<E.

This means...

|x-x0||x2+x0x+x02|<E

I see that some value less than delta shows up here, but don't know what to do from here. I see several other relations that are seemingly useless, such as...

|x|<D+|x0|

Where do I go from here? I know I need to figure out what to pick as my delta, but I don't know how to do this.
• Oct 30th 2012, 08:25 PM
Prove It
Re: Having trouble showing continuity.
You have gotten to \displaystyle \begin{align*} \left| x - x_0 \right| \left| x^2 + x\, x_0 + x_0 ^2 \right| < \epsilon \end{align*}. Great job. From this we have \displaystyle \begin{align*} \left| x - x_0 \right| < \frac{\epsilon}{\left| x^2 + x\, x_0 + x_0^2 \right|} \end{align*}. We need this denominator to be at its maximum so that the fraction on the right will be at its minimum, to give us a value we can set for \displaystyle \begin{align*} \delta \end{align*}. To maximise the denominator, we need an upper bound, so we can use the triangle inequality...

\displaystyle \begin{align*} \left| x^2 + x\,x_0 + x_0^2 \right| &\leq \left|x \right|^2 + \left| x \right| \left|x_0 \right| + \left| x_0 \right|^2 \textrm{ by the Triangle Inequality } \end{align*}

Unfortunately, this denominator is unconstrained, but no matter, we only care about what's happening NEAR \displaystyle \begin{align*} x = x_0 \end{align*}, so we restrict the distance between \displaystyle \begin{align*} x \end{align*} and \displaystyle \begin{align*} x_0 \end{align*} to within some small value, say 1. Then \displaystyle \begin{align*} \left| x - x_0 \right| < 1 \end{align*}.

The first thing we can determine from \displaystyle \begin{align*} \left| x - x_0 \right| < 1 \end{align*} is that

\displaystyle \begin{align*} -1 < x - x_0 &< 1 \\ -1 + x_0 < x &< 1 + x_0 \\ |x| &< 1 + x_0 \end{align*}

So that means \displaystyle \begin{align*} \left| x \right|^2 + \left| x \right| \left|x_0 \right| + \left| x_0 \right|^2 &< \left( 1 + x_0 \right) ^2 + \left( 1 + x_0 \right) \left| x_0 \right| + \left| x_0 \right| ^2 \end{align*}

Therefore we can say \displaystyle \begin{align*} \frac{\epsilon}{\left| x^2 + x\,x_0 + x_0^2 \right|} \end{align*} is minimised at \displaystyle \begin{align*} \frac{\epsilon}{ \left( 1 + x_0 \right)^2 + \left(1 + x_0 \right)\left| x_0 \right| + \left|x _0 \right|^2} \end{align*}.

So finally we can let \displaystyle \begin{align*} \delta = \min \left\{ 1, \frac{\epsilon}{ \left( 1 + x_0 \right)^2 + \left( 1 + x_0 \right) \left| x_0 \right| + \left| x_0 \right|^2 } \right\} \end{align*} and reverse each step to complete your proof.
• Nov 19th 2012, 08:32 PM
wallisonline
Re: Having trouble showing continuity.
What was the reason that Ramses ii built the temple at Abu Simbel?

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