# limit of absolute value

$$\lim_{x \to 0} \frac{\lvert2x-1\rvert - \lvert2x+1\rvert}{x}$$

Defining the function piecewise reveals the limit is in fact, continuous about 0

However when I go to solve it in a normal algebraic manner, the $2x$ terms are canceling, leaving me with an undefined output - 0 in the denominator.

Any hints would be fantastic, I've solved this every way I can think of and I keep getting different answers, none of which are the correct answer.

I did graph this, and it does show a continuous function around $0$ where $f(x) = -4$

My problem here is that I'm being a huge dunce about absolute values.

• While its not a prove, did you graph it? It does give some insight of what's going on at zero. And why exactly do these 2x terms cancel?? – imranfat Oct 19 '14 at 18:49

For $\;x\;$ pretty close to zero, $\;2x-1<0\;,\;\;2x+1>0\;$ , so we have the limit

$$\frac{-2x+1-2x-1}x=\frac{-4x}x=-4\xrightarrow[x\to 0]{}-4$$

• I don't understand why your first 2x term is negative, understanding that is clearly the key to this problem. – James Oct 19 '14 at 19:13
• @JamesGibsonWeber : $$2x-1<0\implies |2x-1|=-(2x-1)=-2x+1$$ The definition of absolute value, and for $\;x\;$ close enough to $\;0\;$ . – Timbuc Oct 19 '14 at 19:17
• Yes. Since $\;x\to 0\;$ you can restrict yourself to, say $\;-\frac14<x<\frac14\;,\;\;x\neq 0\;$ . It doesn't matter whether $\;x\;$ is positive or negative: the absolute values give the same as above. – Timbuc Oct 19 '14 at 19:29
• I think I get it now, thank you – James Oct 19 '14 at 19:30

I recognize this problem from Stewart (or at least something similar). Use the fact that $$|2x-1| = \begin{cases} 2x-1, & 2x-1 > 0 \\ -(2x-1) = 1 - 2x, & 2x-1 < 0 \end{cases}$$ and $$|2x+1| = \begin{cases} 2x+1, & 2x+1 > 0 \\ -(2x+1) = -2x - 1, & 2x+1 < 0\text{.} \end{cases}$$ Examine the behavior around $x = 0$ using these two equations.

# $$\lim_{x \to 0} \frac{|2x-1| - |2x+1|}{x}$$

\begin{align} \frac{|2x-1| - |2x+1|}{x} &= \frac{|2x-1| - |2x+1|}{x} \cdot \frac{|2x-1| + |2x+1|}{|2x-1| + |2x+1|} \\ &= \frac{(2x-1)^2 - (2x+1)^2}{x(|2x-1| + |2x+1|)} \\ &= \frac{-8}{|2x-1| + |2x+1|} \\ \end{align}

and

$$\displaystyle \lim_{x \to 0} \frac{-8}{|2x-1| + |2x+1|} = \frac{-8}{2} = -4$$