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Math Help - Limit with absolute value

  1. #1
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    SOLVED: Limit with absolute value

    \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?
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  2. #2
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    Re: Limit with absolute value

    Nevermind, figured it out.
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  3. #3
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    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.
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  4. #4
    Member sbhatnagar's Avatar
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    Re: Limit with absolute value

    Also it is known that:

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

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

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