Can a straight line be considered a part of a circle? Say we have a circle of infinite radius. If we zoom in infinitely on its perimeter, we should end up looking at a straight line. Intuitively. But, such a line, I believe, should have a curvature of $1/∞$. But a straight line is defined to have $0$ curvature in Euclidean geometry. My interpretation also has problems if we consider parallel lines. Can someone guide me on this?
 A: "circle of infinite radius"
Meaningless.
"If we zoom in infinitely"
Meaningless.
"we should end up looking at a straight line"
A straight line that is a magnification of a single point.  Which is meaningless.
$\infty$ is not a number or a value. It is a concept to describe values that may be arbitrarily large and have no upper bound.
All circles have finite radius.  But for any value $M$ we can have a circle with radius $> M$.  So the radius becomes large the curvature becomes smaller.  We state that as $\lim_{r\to \infty} Curvature (r) = 0$ (and similarly $\lim_{x \to \infty} \frac 1x = 0$).  But that does not mean infinite radius or $0$ curvature or $\frac 1 {\infty}$ ever occur.  They don't and can't.
But yes for a circle with extremely huge radius will have extremely small curvature and the arcs will become arcs that differ from a line segments only very slightly.  But it is never infinite radius or zero curvature.  And although the arc will differ from straight lines only minisculely, they will differ.
A: No, it can't. A straight line can't be considered as part of a circle.
I can remember from my algebra class (long time ago), that at some point we had to take the limit of $x$ to zero. And the teacher was screaming: we never get to zero (my ears are still ringing). To give you an example, for $f(x)=\frac{1}{x}$ we have:
$$
\lim_{x\rightarrow0}f(x)=\lim_{x\rightarrow0}\frac{1}{x}=\infty,
$$
but $f(x)$ does not exist for $x=0$.
