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.
$f(x)$<$g(x)$
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. $\endgroup$