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We define function $f(x) = 5x^3$.

To prove: $f$ is continuous in x, with $x \in \mathbb R.$

Suppose, $\epsilon > 0,$ we choose $\delta = \frac{\epsilon}{5(3x^2 + |3x| +1)}$, then for all $x \in \mathbb R$, with $|x - c| \lt \delta$, we have $|5x^3 - 5c^3| < \epsilon$. We also suppose $\delta < 1.$

|$5x^3 - 5c^3$| = 5|$x^3 - c^3$| = 5|$x - c$||$x^2 + c^2$ + cx| < 5 δ|$x^2 + c^2$ + cx|

|$x^2+c^2+xc$| < |$x^2 + (x + 1)^2+ x(x+1)$| so, |$x^2+c^2+xc$|< $3x^2 + |3x| + 1$

5 δ|$x^2 + c^2$ + cx| < 5 δ ($3x^2 + |3x| + 1$) < ϵ.

Is this proof sufficient?

amWhy
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D.G. van de Schepop
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  • You seem to be trying to use mathjax, and doing quite well. Please note that you can surround entire expressions, within a single pair of dollar signs, so it goes much more quickly. Thanks for the effort! – amWhy Oct 26 '21 at 22:38
  • Click on "Edit" (left bottom). I've formatted a bit to demonstrate full formatting of expressions. And for epsilon, use \epsilon, and for delta, use \delta. Otherwise, keep up formatting efforts! – amWhy Oct 26 '21 at 22:45
  • @amWhy ok thanks for the editing and tips – D.G. van de Schepop Oct 26 '21 at 22:49
  • No problem. Thanks for your efforts here! – amWhy Oct 26 '21 at 22:50
  • Welcome to MSE! <> It looks as if you're proving $f$ is continuous at $c$. If that's right, your choice of $\delta$ can depend on $c$, but not on $x$. (Fixing this may simply be a matter of exchanging a few $x$s with $cs; I haven't checked your calculations carefully.) – Andrew D. Hwang Oct 26 '21 at 23:17
  • Thanks for the warm welcome! Can you explain why I am proofing that f is continuous in c and not in x? – D.G. van de Schepop Oct 26 '21 at 23:46

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No, you can't have your $\delta$ depending on $x$, as you haven't said anything about $x$ yet.

You'll need to 'work backwards' as I've always been taught, begin from what you want to prove to get an expression for $\delta$.

BigManSinceTheDawnOfTime
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