$\lim_{x \to -1} \frac{|x + 1|} { x^2 + 2x + 1}$?

Question

Is the solution: "no limit"

Because:

$$\frac{|x + 1| }{ (x^2 + 2x + 1)} = \frac{1}{(x + 1)}$$

Or is factoring different for $|x + 1|$?

Answer 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 $$ \lim_{x\to-1}\frac{|x+1|}{x^2+2x+1}= \lim_{x\to-1}\frac{1}{|x+1|}=\dots $$