$\lim_{x \to -1} \frac{|x + 1|} { x^2 + 2x + 1}$? Is the solution: "no limit"
Because:
$$\frac{|x + 1| }{ (x^2 + 2x + 1)} = \frac{1}{(x + 1)}$$
Or is factoring different for $|x + 1|$?

You can see that
$$
\frac{|x+1|}{x^2+2x+1}\ne\frac{1}{x+1}
$$
by just plugging in $x=-2$; the left-hand side is
$$
\frac{|-2+1|}{4-4+1}=1
$$
whereas the right-hand side is
$$
\frac{1}{-2+1}=-1
$$
Better, $x^2+2x+1=(x+1)^2=|x+1|^2$, so the correct simplification is
$$
\frac{|x+1|}{x^2+2x+1}=\frac{|x+1|}{|x+1|^2}=\frac{1}{|x+1|}
$$
and so the limit is
$$
\lim_{x\to-1}\frac{|x+1|}{x^2+2x+1}=
\lim_{x\to-1}\frac{1}{|x+1|}=\dots
$$
