How to Proof that the quadratic equation is continuous in the Real numbers?

Question

Intuitively is obvious that an equation of the form $ax^2 +bx+c$ is continuous for all real numbers because it does not have any indeterminacy. But I need a more formal proof

Answer 1

Outline

You can prove the following theorems about continuity:

1. constant is continuous
2. identity is continuous
3. sum of continuous functions is continuous
4. product of continuous functions is continuous

Then just use 1-4 to prove that $ax^2+bx+c$ is continuous.

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1. Constant is continuous

We say that $\forall r \in \Bbb R: \displaystyle \lim_{x \to r} f(x) = f(r)$ where $f(x) = C$ for a constant $C$.

That is, $\forall r \in \Bbb R: \forall \varepsilon > 0 : \exists \delta > 0: \forall x \in \Bbb R: |x-r| < \delta \implies |f(x)-f(r)| < \varepsilon$.

Letting $\delta = 1$ solves the problem.

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2. Identity is continuous

We say that $\forall r \in \Bbb R: \displaystyle \lim_{x \to r} f(x) = f(r)$ where $f(x) = x$ for all real number $x$.

That is, $\forall r \in \Bbb R: \forall \varepsilon > 0 : \exists \delta > 0: \forall x \in \Bbb R: |x-r| < \delta \implies |f(x)-f(r)| < \varepsilon$.

Letting $\delta = \varepsilon$ solves the problem.

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3. Sum of continuous functions is continuous

We say that $\forall r \in \Bbb R: \displaystyle \lim_{x \to r} f(x) + g(x) = f(r) + g(r)$ where $f$ and $g$ are continuous functions.

That is, $\forall r \in \Bbb R: \forall \varepsilon > 0 : \exists \delta > 0: \forall x \in \Bbb R: |x-r| < \delta \implies |f(x)+g(x)-f(r)-g(r)| < \varepsilon$.

Plug in $\dfrac\varepsilon2$ to the expressions of $f$ and $g$ to obtain $\delta_1$ and $\delta_2$ respectively, then take the minimum.

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4. Product of continuous functions is continuous

We say that $\forall r \in \Bbb R: \displaystyle \lim_{x \to r} f(x) g(x) = f(r) g(r)$ where $f$ and $g$ are continuous functions.

That is, $\forall r \in \Bbb R: \forall \varepsilon > 0 : \exists \delta > 0: \forall x \in \Bbb R: |x-r| < \delta \implies |f(x) g(x) - f(r) g(r)| < \varepsilon$.

This is left to the reader as an exercise.