Evaluate limit of a absolute value

**johnny12345** · October 27th 2010, 04:04 PM

lim as x goes to 0

(abs(2x-1)-abs(2x+1))/x

Something with piecewise?

**chisigma** · October 28th 2010, 04:55 AM

For $0<x<\frac{1}{2}$ is $|2x-1|= 1-2x$ and $|2x+1|= 2x+1$ so that is...

$\displaystyle \lim_{x \rightarrow 0+} \frac {|2x-1|- |2x+1|}{x} = \lim_{x \rightarrow 0+} \frac {-4x}{x} = -4$ (1)

For $-\frac{1}{2}<x<0$ is [also] $|2x-1|= 1-2x$ and $|2x+1|= 2x+1$ so that is...

$\displaystyle \lim_{x \rightarrow 0-} \frac {|2x-1|- |2x+1|}{x} = \lim_{x \rightarrow 0-} \frac {-4x}{x} = -4$ (2)

Kind regards

$\chi$ $\sigma$

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