Question

Trouble!
lim x->-2 (abs(x^2+2x))/(2x+4)

Answer 1

sorry \[\lim_{x \rightarrow -2^{-}}\]

Answer 2

Factor the numerator and factor the denominator first.

Answer 3

How would I go about that with the absolute values?

Answer 4

\[
\lim_{x \rightarrow -2^{-}} \frac{|x^2+2x|}{2x+4} =
\lim_{x \rightarrow -2^{-}} \frac{|x(x+2)|}{2(x+2)}
\]

Answer 5

Alright makes sense

Answer 6

When x approaches -2 from the left of -2, what will the numerator be without the absolute bars?

Answer 7

\[0^{-}\]

Answer 8

When x approaches -2 from the left, x < -2.

So the x within the absolute bar can be replaced by -x without the absolute bar as both will yield +2 when x->-2.
Similarly, (x+2) will be negative when x < -2, so (x+2) within the absolute bars can be replaced by -(x+2) without the absolute bars.
\[
\lim_{x \rightarrow -2^{-}} \frac{|x(x+2)}{2(x+2)} =
\lim_{x \rightarrow -2^{-}} \frac{-x * -(x+2)}{2(x+2)} =
\lim_{x \rightarrow -2^{-}} \frac{-x * -1}{2} = -1
\]

Answer 9

Ohhh, okay. I was getting confused with the absolute values and limits. I get it now though thanks!

Answer 10

You are welcome.