# Limit of functions absolute value

\begin{align} &\lim\limits_{x\to0} \frac{|3x-1|-|3x+1|}x\\ =&\lim\limits_{x\to0} \frac{(3x-1)^2-(3x+1)^2}{x(|3x-1|+|3x+1|)}\\ =&\lim\limits_{x\to0} \frac{-12x}{x(|3x-1|+|3x+1|)} = \frac{-12}{1+1}=-6 \end{align}

I have a hard time understanding how this is possible.

First it looks like they are multiplying numerator and denominator with the same,|3x-1|+|3x+1|, I am do not get how this converts to $(3x-1)^2-(3x+1)^2$,

Second, in line 3, in the denominator it says x(....), when we substitute 0 for x, why are we not multiplying everything within the ().

I first thought this had no limit, but clearly I was wrong.

• The formula $x^2-y^2 = (x+y)(x-y)$ is used with $x = \lvert 3x-1\rvert, y = \lvert 3x+1\rvert$. – Gaussler Jul 6 '14 at 6:07
• $(|a|-|b|)*(|a|+|b|)=(|a|)^2+(|a||b|)-(|b||a|)-(|b|)^2=(|a|)^2-(|b|)^2=a^2-b^2$ – Vikram Jul 6 '14 at 6:13

$a^2-b^2=(a-b)(a+b)$ for any real numbers $a$ and $b$. Just multiply to see the proof. Substitute a and b with $|3x-1|+|3x+1|$ , $|3x-1|-|3x+1|$, respectively. For your second question, $x$ cancels both at the denominator and numerator.