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Math Help - Absolut value of x

  1. #1
    Junior Member
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    Absolut value of x

    Hi everyone.
    I'm having a smole problem with this function H(x)=(2+|x|)/(1-|x|) . So my function has 4 equalities :
    H(x)=(2+x)/(1-x) for x>0,
    H(x)=(2+x)/(1+x) for x>0,
    H(x)=(2-x)/(1-x) for x<0,
    H(x)=(2-x)/(1+x) for x<0.

    I need to find lim x->1+ H(x) - so 1 from right side.
    Shall i look on H(x)=(2+x)/(1-x) or maybe H(x)=(2-x)/(1-x) ???? for both of them lim is -infinity.

    Also lim x-> -1 H(x)
    and again H(x)=(2+x)/(1+x) or maybe H(x)=(2-x)/(1+x) . For the first one lim is 3/2 for the second one lim is 1/2.


    Also lim -> infinity H(x)
    and lim ->0 H(x)

    I'll be thankful for any help!
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  2. #2
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    Quote Originally Posted by Snowboarder View Post
    Hi everyone.
    I'm having a smole problem with this function H(x)=(2+|x|)/(1-|x|) . So my function has 4 equalities :
    H(x)=(2+x)/(1-x) for x>0,
    H(x)=(2+x)/(1+x) for x>0,
    H(x)=(2-x)/(1-x) for x<0,
    H(x)=(2-x)/(1+x) for x<0.
    The four parts of your graph are

    H(x)=\frac{2-x}{1+x}\ x<-1

    H(x)=\frac{2-x}{1+x}\ -1<x\le0

    H(x)=\frac{2+x}{1-x}\ 0\le x<1

    H(x)=\frac{2+x}{1-x}\ x>1

    When x>1, H(x) is negative. Therefore \lim_{x\,\to\,1^+}{H(x)}=-\infty.

    The curve is symmetrical about the y-axis, so \lim_{x\,\to\,-1^-}{H(x)}=\lim_{x\,\to\,1^+}{H(x)}=-\infty.

    \lim_{x\,\to\,\infty}{H(x)}=\lim_{x\,\to\,\infty}{  \frac{2+x}{1-x}}=\lim_{x\,\to\,\infty}{\frac{\frac{2}{x}+1}{\fr  ac{1}{x}-1}}=-1

    And H is continuous at 0, so \lim_{x\,\to\,0}{H(x)}=\frac{2+|0|}{1-|0|}=2.
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