# Thread: Absolut value of x

1. ## 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!

2. Originally Posted by Snowboarder
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

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

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

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

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

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

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

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

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