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I need to show that a function $g:V\to\mathbb{R}$ given by $g(x)=\frac{a(x)}{b(x)}$ is continuous, where $a(x)=x^2-3x+2$ and $b(x)=x^2-4x+3$. I know $V=\mathbb{R}\setminus\{1,3\}$

My guess is that I need to use this definition: $$\forall\epsilon>0\ \exists N\in\mathbb{N}:n \geq N \implies |f(x)-f(x_n)|<\epsilon$$ My guess is that I need to use the definition above to prove $a(x)$ and $b(x)$ are continues and the use this proposition to argue for $g(x)$ is continues: $$(f/g):V\to\mathbb{R}\quad \text{given by}\quad (f/g)(x)=f(x)/g(x)$$ is continuous and $g(x)\neq0$

Overall am I a bit lost to how I show that $g(x)$ is continues. I would apricate any help:)

CiaPan
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luja
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    Do you have any theorems/results that you can use? Such as "if $f,g$ are continuous, and $g\neq 0$, then $\frac fg$ is continuous", or do you need to show this as well? – DominikS Nov 03 '23 at 10:58
  • I don't need to prove that as well. I have the result that states what you mentioned. – luja Nov 03 '23 at 11:02
  • Similarly, are you allowed to use a result that says that the sum or product of continuous functions is continuous again? – DominikS Nov 03 '23 at 11:04
  • Yes I am. I am allowed to use the results that says the addition of of continuous functions is continuous. The product of of continuous functions is continuous and the one you mentioned in your first comment. – luja Nov 03 '23 at 11:08
  • Okay, so maybe you don't need to apply the $\epsilon$-$\delta$-criterion at all? In the case that you can use all the above, this will probably answer your question. – DominikS Nov 03 '23 at 11:10
  • I will take a look. Thank you:) – luja Nov 03 '23 at 11:11
  • You'll have 2 steps: (1) the polynomials are sums and products of continuous functions ($x$ and $c\in\mathbb R$, i.e. constants, both are trivially continuous), and (2) the quotient of two polynomials will be continuous due to the result we discussed in the comments above. – DominikS Nov 03 '23 at 11:16
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    I believe I still need to use the $\epsilon$-$delta$-criterion to show that $a(x)$ and $b(x)$ are continuous. – luja Nov 03 '23 at 11:16
  • You might find this interesting, but maybe to general for your application. – DominikS Nov 03 '23 at 11:22
  • Also, this will give you a very good explanation of how to show continuity of a polynomial, using a $\epsilon$-$\delta$ criterion. – DominikS Nov 03 '23 at 17:11

1 Answers1

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I will give a sketch of the main idea.

We want to show continuity of $b(x)$ at some $x=x_0$. Pick an arbitrary $\epsilon > 0$, and keep it fixed in the following. Next, we evaluate (for arbitrary $\Delta x\in\mathbb R$): $$b(x_0 + \Delta x) = b(x_0) +\Delta x(2x_0 - 4 +\Delta x).$$ Given our choice of $x_0$ and $\epsilon$, we need to find a $\delta >0$ such that $|\Delta x(2x_0 - 4 +\Delta x)|<\epsilon$ for all $|\Delta x|<\delta$. If we can find such a $\delta$, this would prove continuity at $x_0$ (note that you need to be able to find such a $\delta$ for every choice of $\epsilon > 0$).

The following is a heuristic to convey the main idea of what remains to be done. For small enough $\Delta x$ and $2x_0 - 4$ not close to zero, we would expect something like this: $$|\Delta x(2x_0 - 4 +\color{red}{\Delta x})|<\epsilon\quad \Leftrightarrow\quad |\Delta x| < \frac{\epsilon}{|2x_0 - 4+\color{red}{\Delta x}|}\approx \frac{\epsilon}{|2x_0 - 4|},$$ so you might choose $\delta \approx \frac{\epsilon}{2x_0 - 4}$ (or a bit smaller, but we are not trying to be rigorous yet). Having found such a $\delta$ for every $\epsilon$, this would complete the proof. However, $\ldots$.

$\ldots$ to be rigorous, you need to solve the inequality $|\Delta x(2x_0 - 4 +\Delta x)|<\epsilon$ for $\Delta x$ and prove that for any choice of $\epsilon > 0 $ and $x_0$, you can find corresponding $\delta_L <0 < \delta_R$ such that the inequality is satisfied for all $\Delta x \in (\delta_L, \delta_R)$. You would pick $\delta=\min(-\delta_L, \delta_R)$, or anything smaller. Often, $\delta_{L,R}$ will be the appropriate roots of the polynomial(s) $\pm y(2x_0 - 4 + y) - \epsilon$, but in general you will need to

  • take the absolute value into account (i.e. it's not simply a purely quadratic inequality), and
  • be able to argue that all $\Delta x \in (\delta_L, \delta_R)$ satisfy the inequality (and not just $\Delta x =\delta_L$ and $\Delta x = \delta_R$).

I hope this sheds some light on your question. But this should also explain why the theorems we discussed in the comments are so helpful - it is very tedious to show continuity of even a simple thing as a quadratic polynomial rigorously.

DominikS
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