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I know there are a lot of answers regarding continuity of polynomials. But, this question is different.

We need to have $ \lim_{x\to a} {x^n} = a^n$ , $n \in N$ , to be able to prove that polynomials are continuous. This fact is derived from the product rule (or may be it can't be, which is my question). The product rule is proved using square roots, so it assumes the existence of square roots. The fact that For every non negative number $x$, it's $n^{th}$ root exists, i.e. $x^n$ is invertible, assumes the continuity of $x^n$ because this is proven using Intermediate value Theorem.
Bam - Circular reasoning ! Or am I Wrong ? Here's the only proof of product rule which I know : Assume $ \lim_{x\to a} {f(x)} = L$ and $ \lim_{x\to a} {g(x)} = K$

Let $ϵ > 0$ be any positive number Hence, $∃\delta_1> 0 ∶ 0<|x-a|<δ_1⟹|f(x)-L|<\sqrt{\epsilon}$

And $∃δ_2>0 ∶ 0<|x-a|<δ_2 ⟹ |g(x)-K|<\sqrt{\epsilon}$

Let $δ=\min\{δ_1,δ_2\}$

Hence, $0<|x-a|<δ$

$|(f(x)-L)(g(x)-K)-0|<\sqrt{\epsilon} \sqrt{\epsilon} = ϵ$

Hence, $\lim_{x \to a} {(f(x)-L)(g(x)-K)} = 0$

$\lim_{x \to a} {(f(x)g(x)-Kf(x)-Lg(x)+KL)} = 0$

And then the result follows. Even if we use $\epsilon$ in place of $√ϵ$ , we end up with $0<|x-a|<δ⟹|(f(x)-L)(g(x)-K)-0|<\epsilon\cdot \epsilon = ϵ^2$ , and then we have to prove that the range of $\epsilon^2$ is $[0,\infty]$, which amounts to proving that for each number in $[0,\infty]$ , a corresponding square root exists.

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    $\begingroup$ You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $\lim_{x\to a}$ and $Lim_{x\to a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit. $\endgroup$ – Martin Sleziak Jan 7 '19 at 8:57
  • $\begingroup$ Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake. $\endgroup$ – Steve Jan 7 '19 at 9:01
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    $\begingroup$ Why don't you use the fact that $$|f(x) g(x) - LK|\leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $\epsilon /2$? $\endgroup$ – Paramanand Singh Jan 7 '19 at 9:10
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    $\begingroup$ Use the fact that $g(x) \to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<\epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<\epsilon/(2(|L|+1))$ to handle second term. $\endgroup$ – Paramanand Singh Jan 7 '19 at 9:29
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    $\begingroup$ In general when you are dealing with $\epsilon, \delta$ proofs the expression based on these $\epsilon, \delta$ should not involve operations other than $+, -, \times, /$. $\endgroup$ – Paramanand Singh Jan 7 '19 at 9:32
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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

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

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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 $\epsilon^\prime$.

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  • $\begingroup$ IOW, you don't need the fact that $x\mapsto x^2$ is continuous, just that it's monotonic for $x>0$. $\endgroup$ – leftaroundabout Jan 7 '19 at 20:59

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