Confusion regarding taking the square root given an absolute value condition. From the Generating function for Legendre Polynomials:
$$\Phi(x,h)=(1-2xh+h^2)^{-1/2}\quad\text{for}\quad \mid{h}\,\mid\,\lt 1$$
My text states that: 

For $x=1$ $$\Phi(1,h)=\color{red}{(1-2h+h^2)^{-1/2}=\frac{1}{1-h}}=1+h+h^2+\cdots$$

My question is about the justification of the equality marked $\color{red}{\mathrm{red}}$. 
Since although
$$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((1-h)(1-h)\Big)^{-1/2}$$$$=\Big((1-h)^2\Big)^{-1/2}=\color{#180}{\frac{1}{1-h}}=1+h+h^2+\cdots\tag{1}$$
as required.
I could also write $$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((h-1)(h-1)\Big)^{-1/2}$$$$=\Big((h-1)^2\Big)^{-1/2}=\color{blue}{\frac{1}{h-1}}=\frac{1}{h}\left(1-\frac{1}{h}\right)^{-1}\ne 1+h+h^2+\cdots\tag{2}$$
Why is it that $\Phi(1,h)$ is equal to $(1)$ but not equal to $(2)$? I think the answer lies somewhere with the fact that $\bbox[yellow]{\mid \,h\mid\,\lt 1}$ but I'm not quite sure how to justify it completely. 
You may be wondering why I'm asking such a question, as one user pointed out that given my rep I should know this answer easily. The fact of the matter is I do not know the answer. I'm just looking for a simple English explanation as to why $\Phi(1,h)\ne \color{blue}{\dfrac{1}{h-1}}$. Although I usually prefer hints; I was given so many hints on this question but none of them helped. So if someone could literally spell out the answer to me I would be most grateful.
Thank you.
 A: Of course for real $x$, we have $\big(x^2\big)^{-1/2} = \,\mid x \,\mid^{-1}$ using the principal square root.  In your case, it is assumed that $\mid \, h\mid \, \lt 1$, which tells you
$$
\left|\frac{1}{1-h}\right| = \frac{1}{1-h}
$$
but
$$
\left|{\frac{1}{h-1}}\right| = -{\frac{1}{h-1}}
$$
A: For $$\mid\, h \mid\lt 1$$ 
$$\implies -1\lt h \lt 1 $$
$$\implies 1\gt h \gt -1$$
$$\implies 1-h\gt h-h\gt 0$$
Since $$1-h\gt 0$$
$$\implies \frac{1}{1-h}\gt 0$$
$$\implies \left|\frac{1}{1-h}\right| = \bbox[#AFF]{\frac{1}{1-h}}$$

For $$\mid\, h \mid\lt 1$$ 
$$\implies -1\lt h \lt 1 $$
$$\implies h-1\lt 1-1 \lt 0$$
Since $$h-1\lt 0$$
$$\implies \frac{1}{h-1}\lt 0$$
$$\implies \left|{\frac{1}{h-1}}\right| = -{\frac{1}{h-1}} = \bbox[#AFF]{\frac{1}{1-h}}$$
as before.

So $$\Phi(1,h)=\underbrace{\Big((1-h)^2\Big)^{-1/2}=\left|\frac{1}{1-h}\right|}_{\color{red}{\Large\text{Using}\, \big(x^2\big)^{-1/2}\, = |x|^{-1}}}=\bbox[#AFA]{\frac{1}{1-h}}=1+h+h^2+\cdots$$

Also $$\Phi(1,h)=\underbrace{\Big((h-1)^2\Big)^{-1/2}=\left|\frac{1}{h-1}\right|}_{\color{red}{\Large\text{Using}\, \big(x^2\big)^{-1/2} =\, |x|^{-1}}}=\underbrace{-\frac{1}{h-1}=\bbox[#AFA]{\frac{1}{1-h}}}_{\large\color{#F80}{\text{By the argument shown above}}}=1+h+h^2+\cdots$$

So to summarize; writing $$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((1-h)(1-h)\Big)^{-1/2}=\Big((1-h)^2\Big)^{-1/2}$$ 
or 
$$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((h-1)(h-1)\Big)^{-1/2}=\Big((h-1)^2\Big)^{-1/2}$$ both result in the same answer of 
$$\bbox[yellow]{\frac{1}{1-h}}$$ as they should do. 

A special thanks goes to @GEdgar and @A.S. for giving me the knowledge necessary to make this answer.
