Having trouble showing continuity.

I wish to show f(x)=x^{3} such that x is an element of the real numbers is continuous at an arbitrary x_{0}.

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-x_{0}|<D implies |f(x)-f(x_{0})|<E

So, we have...

|x-x_{0}|<D should imply |x^{3}-x_{0}^{3}|<E.

This means...

|x-x_{0}||x^{2}+x_{0}x+x_{0}^{2}|<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+|x_{0}|

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.

Re: Having trouble showing continuity.

You have gotten to . Great job. From this we have . 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 . To maximise the denominator, we need an upper bound, so we can use the triangle inequality...

Unfortunately, this denominator is unconstrained, but no matter, we only care about what's happening NEAR , so we restrict the distance between and to within some small value, say 1. Then .

The first thing we can determine from is that

So that means

Therefore we can say is minimised at .

So finally we can let and reverse each step to complete your proof.

Re: Having trouble showing continuity.

What was the reason that Ramses ii built the temple at Abu Simbel?

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