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

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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.

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

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  • $\begingroup$ Did see the first one but didnt think that it worked for absolute values as well, that is great. And the cancel of x values, I hope was a result of being on my last hour of 10 hour night shift:P $\endgroup$ – ALEXANDER Jul 6 '14 at 6:44
  • $\begingroup$ Btw, I am looking in the tekstbook but I see that there are other cases where it does not exist a limit, how do one know that there is no arithmetic that can be done, such that it is possible to find a limit. $\endgroup$ – ALEXANDER Jul 6 '14 at 6:51

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