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Let $f_n \to \Bbb R$ be defined by $ f_n= \frac{8n^3+4n^2+2}{2n^3+11n+7}x^2 $

Prove $f_n$ is continuous:

let $ \epsilon\gt0 $ be given

let $a \in \Bbb R$ be given

select $ \delta \gt 0 $ such that $ \delta = min {( 1, \frac{\epsilon}{\frac{8n^3+4n^2+2}{2n^3+11n+7}(1+2|a|)}}) $

then for all $x\in \Bbb R $ with $|x-a|\lt\delta $ wehave:

$$|f(x)-f(a)|=$$

$$|\frac{8n^3+4n^2+2}{2n^3+11n+7}x^2-\frac{8n^3+4n^2+2}{2n^3+11n+7}a^2|$$

$$ |\frac{8n^3+4n^2+2}{2n^3+11n+7}|*|x^2-a^2|$$

$$ |\frac{8n^3+4n^2+2}{2n^3+11n+7}||x+a||x-a|$$

since triangular inequality $$\leq|\frac{8n^3+4n^2+2}{2n^3+11n+7}|(|x|+|a|)|x-a| $$

since $|x|\lt|\delta +a|$ $$\lt |\frac{8n^3+4n^2+2}{2n^3+11n+7}|(\delta+2|a|)|x-a| $$

since $\delta \lt 1 $ $$\lt|\frac{8n^3+4n^2+2}{2n^3+11n+7}|(1+2|a|)|x-a| $$

since $ |x-a|\lt\delta $ $$\lt |\frac{8n^3+4n^2+2}{2n^3+11n+7}|(1+2|a|)\delta $$

$$\leq |\frac{8n^3+4n^2+2}{2n^3+11n+7}|(1+2|a|) \frac{\epsilon}{|\frac{8n^3+4n^2+2}{2n^3+11n+7}|(1+2|a|)}$$

$$ =\epsilon $$

I think there might be other ways to do it, but please let me know if I made any mistakes and I would hear your opinion on how I solved it and how you would solve it.

Shervan
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    It's called "continuously differentiable" – amsmath Oct 30 '19 at 12:45
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    Welcome to Mathematics Stack Exchange. I would just say $f_n(x)$ is a polynomial function and therefore continuous – J. W. Tanner Oct 30 '19 at 12:49
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    I would just say that most of your proof is really showing that $x\mapsto x^2$ is continuous, with some extras. Unless you are specifically told to do so from $\varepsilon-\delta$ definition, it is very inefficient. Can you use that product of continuous functions is continuous? – Ennar Oct 30 '19 at 13:30
  • I was told to use the epsilon,delta definition, but you are right, I should've just put a constant in front of the function and prove its continuity. thanks! – Shervan Oct 30 '19 at 14:21

2 Answers2

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Your proof is fine and nicely written. It would have been much easier though to have said at the outset 'Let $N$ be the constant $\frac{8n^3+4n^2+2}{2n^3+11n+7}$'.

  • I thought about that, but the thing is I have a professor who wants everything to be perfect... so I was in doubt. Thanks!! – Shervan Oct 30 '19 at 13:01
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    Making good choices of constants/variables is all a part of making 'perfect' proofs so I'm sure your Prof would approve of simplifying constants when appropriate. –  Oct 30 '19 at 13:14
  • You are right. Thanks. I would try it again later with a constant. – Shervan Oct 30 '19 at 14:23
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I would leverage the theorems that describe the continuity of various functions to conclude...

$ f_n= \frac{8n^3+4n^2+2}{2n^3+11n+7}x^2 $ is a rational function, so it must be continuous on $\mathbb{R}$ except for all $n$ such that $2n^3+11n+7 = 0$.

RyRy the Fly Guy
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