Need Help : Proving polynomials are continuous, without circular reasoning

Solution 1:

The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < \epsilon < 1$, in which case $\epsilon^2 < \epsilon$. Or you could use a convergent sequence $x_n$ such that $\displaystyle \lim_{n \mathop \to \infty} x_n = a$, and show that

$$\lim_{n \to \infty}f(x_n)g(x_n) = f\left( \lim_{n \to \infty} x_n \right)g\left(\lim_{n \to \infty} x_n\right) = f(a)g(a)$$

Solution 2:

You don't need existence of square root. You just need to have that $\forall \epsilon > 0 \ \exists \ \epsilon^\prime > 0$ s. t. ${\epsilon^\prime}^2 < \epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {\epsilon^\prime}^2 < \epsilon$ which gives you what you need anyway. You don't need the bound $\epsilon$ to actually be reachable.

For $\epsilon < 1$, $\epsilon$ itself can serve as $\epsilon^\prime$.