Circle is a straight line I was thinking about regular polygons when I tried this out.
The sum of the internal angles of an $n$ - sided polygon is $\frac{180(n-2)}{n}$. The limit of this function as n approaches infinity is $180$. Since a circle has infinite sides, the internal angles of the circle are $180$ degrees, which is a straight line.
Is there a flaw in my reasoning? If so, what is it?
 A: To summarise, the flaw is that you're treating infinity as if it's something finite. You can't throw around infinities and infinitesimals too casually. My answer will have a few hand-wavy arguments, but it should be enough for you to realise why your argument was flawed.
In your argument you're taking the limit as side-lengths ($l$) go to zero and the limit as the number of sides ($n$) tend to infinity simultaneously. The rule for "simultaneous limits" is that there must be no dependence on order. This means you want:
\begin{align*}
\lim_{n\rightarrow\infty}\lim_{l\rightarrow0}P(n, l) = \lim_{l\rightarrow 0}\lim_{n\rightarrow\infty}P(n, l) = \mathrm{Circle}
\end{align*}
Where $P(n, l)$ is a regular polygon with side-lengths $l$, and $n$ sides.
Imagine you have an infinite number of unit-length line segments. If you use $3$ of them, you have an equilateral triangle with side lengths of $1$. If you use $n$ of them, you get the regular $n$ sided polygon with side lengths of $1$. Suppose you were to measure the interior angle of such a polygon. If this angle is less than $180^{\circ}$, then you can work out how many sides were used to construct it. In other words, it's still only a finite number of sides. Thus $\lim_{n\rightarrow\infty}P(n, l)$ must be equal to an infinite straight line. This no longer has any dependence on $l$, because $\infty/l = \infty$ for any $l\neq 0$. Therefore,
\begin{align*}
\lim_{l\rightarrow 0}\lim_{n\rightarrow\infty}P(n, l) = \lim_{l\rightarrow 0}(\lim_{n\rightarrow\infty}P(n, l)) = \lim_{l\rightarrow 0}(\mathrm{Infinite\ straight\ line}) = \mathrm{Infinite\ straight\ line}
\end{align*}
How about the other limit? Suppose you have chosen some number of sides, and their lengths can vary. If you have $3$ sides and they have lengths of $1$, you have an equilateral triangle with side lengths of $1$. For any finite $l$, you will have equilateral triangle with side length $l$. Now if you suppose you have a tiny tiny polygon and it's still a triangle, the side length is clearly some finite $l > 0$. Hence $\lim_{l\rightarrow0}P(n, l)$ must be a single point. No matter how many "sides" you give to a single point it will always just be a single point. Therefore,
\begin{align*}
\lim_{n\rightarrow\infty}\lim_{l\rightarrow0}P(n, l) = \lim_{n\rightarrow\infty}(\lim_{l\rightarrow0}P(n, l)) = \lim_{n\rightarrow\infty}(\mathrm{Single\ point}) = \mathrm{Single\ point}
\end{align*}
Clearly a single point is not an infinite line, and they're both certainly not circles. The flaw in your argument is that "the polygon formed when side lengths go to zero and the number of lines goes to infinity" cannot be uniquely described as a circle.
Also I apologise to everybody for my loose use of the word "polygon" 
