# Thread: Limit with absolute value

1. ## SOLVED: Limit with absolute value

$\displaystyle \lim_{x\rightarrow 0} \frac{|3x-1|-|3x+1|}{x}$

I argue that the nominator goes to 0 and we get indeterminate form 0/0. But WolframAlpha says it is supposed to be -6. How?

2. ## Re: Limit with absolute value

Nevermind, figured it out.

3. ## Re: Limit with absolute value

Great. For the benefit of others who may wonder about this, note that, if x is close to 0, 3x-1 will be negative and 3x+1 will be positive. That is, |3x-1|= -(3x-1)= -3x+ 1 and |3x+1|= 3x+1. |3x-1|- |3x+1|= -3x+1- 3x-1= -6x. -6x/x= -6.

You are correct that just taking x= 0 gives the indeterminate form 0/0 but that does not contradict the fact that that the limit is -6.

4. ## Re: Limit with absolute value

Also it is known that:

$\displaystyle \frac{d|x|}{dx}=\frac{x}{|x|}$

On the use of L Hopital's rule, $\displaystyle \lim_{x \to 0}\frac{|3x-1|-|3x+1|}{x}$ will become :

$\displaystyle \lim_{x \to 0}(\frac{9x-3}{|3x-1|}-\frac{9x+3}{|3x+1|})=-6$