TITLE: Problem with absolute value
QUESTION [0 upvotes]: Say that $|\sqrt{x}-1| < \epsilon$. I am having a problem with handling this inequality. I want to exclude x.
I. 
$|\sqrt{x}-1| < \epsilon$
$|\sqrt{x}| - |1| \leq |\sqrt{x}-1| < \epsilon$
$|\sqrt{x}| - |1| < \epsilon$  
$|\sqrt{x}| < \epsilon + |1|$
$|\sqrt{x}| < \epsilon + 1$
$-(\epsilon + 1)< \sqrt{x} < \epsilon + 1$
$(-(\epsilon + 1))^2< x < (\epsilon + 1)^2$
If however I try this:
II.
$|\sqrt{x}-1| < \epsilon$
$-\epsilon < \sqrt{x}-1 < \epsilon$
$1 - \epsilon < \sqrt{x} < 1+ \epsilon$
$(1-\epsilon)^2 < x < (1+\epsilon)^2$
What am I doing wrong in I. or II.?

REPLY [0 votes]: In $I$, the first step is incorrect :
$$
|\sqrt{x} - 1| <\epsilon
$$
is not equivalent to
$$
|\sqrt{x}| - 1 < \epsilon
$$
$II$ is correct though.
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