Thread: 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.

You have gotten to $\displaystyle \displaystyle \begin{align*} \left| x - x_0 \right| \left| x^2 + x\, x_0 + x_0 ^2 \right| < \epsilon \end{align*}$. Great job. From this we have $\displaystyle \displaystyle \begin{align*} \left| x - x_0 \right| < \frac{\epsilon}{\left| x^2 + x\, x_0 + x_0^2 \right|} \end{align*}$. 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 $\displaystyle \displaystyle \begin{align*} \delta \end{align*}$. To maximise the denominator, we need an upper bound, so we can use the triangle inequality...

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

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

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

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


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

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


So finally we can let $\displaystyle \displaystyle \begin{align*} \delta = \min \left\{ 1, \frac{\epsilon}{ \left( 1 + x_0 \right)^2 + \left( 1 + x_0 \right) \left| x_0 \right| + \left| x_0 \right|^2 } \right\} \end{align*}$ and reverse each step to complete your proof.

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