$\vert(1/\sqrt{x}-3) +1\vert$ Having trouble getting this to $\vert x-4\vert$ $$|\frac{1}{\sqrt{x}-3}=-1|$$
I need to somehow move from the above function and arrive at $| x - 4 |$
This is part of a limits problem (an introductory one), but it's the algebra that's stumping me... I've checked out the problem and it's correct, so there must be a way and I am totally missing it. Please help!
(also, please forgive my English, I may not be using the proper math terminology)
To clarify:
I have a limit:
$$
\lim_{x\rightarrow 4} \frac{1}{\sqrt{x}-3}=-1
$$
and I am asked to prove this is correct by following this:
$$
0 < |x-4| < \delta \rightarrow |\frac{1}{\sqrt{x}-3} - (-1) | < \epsilon
$$
As explained in class, I am supposed to take the function that has the square root involved and find a way to make it match with what's on the left of the arrow. 
 A: You are I think looking at 
$$\left|\frac{1}{\sqrt{x}-3}+1\right|.$$
Bring this to a common denominator.
We get
$$\left|\frac{\sqrt{x}-2}{\sqrt{x}-3}\right|.$$
Multiply top and bottom by $\sqrt{x}+2$. We get
$$\frac{|x-4|}{|\sqrt{x}-3|(\sqrt{x}+2)}.$$
Remark: We started with absolute value signs around the expression, since you put some stress on wanting that. But it is at least more pleasant to type if we do the algebra on without absolute values, and take absolute values at the end. 
Added: Presumably the purpose of the transformation is to prove that the limit of our expression as $x\to 4$ is $0$. So suppose we are given an $\epsilon \gt 0$. We want to find a $\delta$ such that if $|x-4|\lt \delta$ then our expression has absolute value less than $\epsilon. 
As a first step, we make sure that $\delta \lt 1/2$. The purpose of this is to make sure $|\sqrt{x}-3|$ is not too small. That puts $x$ between $3.5$ and $4.5$, making $\sqrt{x}\lt \sqrt{4.5}\approx 2.12132\lt 2.2$. It follows that $|\sqrt{x}-3|\gt 0.8$.  
The $\sqrt{x}+2$ part in the denominator is $\gt 2$. So if $\delta\lt 1/2$, the absolute value of the denominator is greater than $1.6$. (We could do better, but don't need to bother. Worse is fine too, as long as we get some bound.)
It follows that as long as we ensure that $\delta\lt \frac{1}{2}$, we have
$$\left|\frac{1}{\sqrt{x}-3}+1\right|\lt \frac{|x-4|}{1.6}.$$
To make this less than $\epsilon$, we take $\delta=(1.6)\epsilon$. Not quite! We make $\delta=\min(1/2,(1.6)\epsilon)$, because our inequality was obtained under the assumption that $\delta\lt 1/2$.
