Is it possible to find a McLaurin Series for the following function? Since the limit as $y$ approaches $0$ is $\infty$ for the function $g(y$) = $\frac{1}{\sqrt{1+y^2} - \sqrt{1-y^2}}$, can we say the MacLaurin Series in this case does not exist? And if this is case, what methods are there for approximating the function near $0$?
 A: Since,
as Mark Viola showed,
$f(y)
=\dfrac{1}{\sqrt{1+y^2}-\sqrt{1-y^2}}
=\dfrac{\sqrt{1+y^2}+\sqrt{1-y^2}}{2y^2}
$,
you can approximate the function
for small $y$ 
($y$ away from $0$ is no problem).
We can use the generalized binomial theorem
in the special case
$\sqrt{1+x}
=\sum_{n=0}^{\infty} \binom{1/2}{n}x^n
=\sum_{n=0}^{\infty} \binom{2n}{n}\dfrac{(-1)^{n+1}x^n}{4^n(2n-1)}
=1-\frac12 x-\frac18 x^2+...
$.
For small $y$,
$f(y)
\approx \dfrac{(1+\frac12 y^2-\frac18 y^4)+(1-\frac12 y^2-\frac18 y^4)}{2y^2}
= \dfrac{2-\frac14 y^4}{2y^2}
= \dfrac{1}{y^2}-\frac18 y^2
$.
You can take as 
many terms as you want:
$\begin{array}\\
\sqrt{1+y^2}+\sqrt{1-y^2}
&=\sum_{n=0}^{\infty}  \binom{1/2}{n}((y^{2n}+(-1)^ny^{2n})\\
&=\sum_{n=0}^{\infty}  \binom{1/2}{2n}2y^{4n}\\
&=2(1-\frac18 y^4 - \frac{5}{128}y^6+...)\\
\end{array}
$
so
$f(y)
=\frac1{y^2}-\frac18 y^2 - \frac{5}{128}y^4+...
$.
A: First, we can write
$$\frac{1}{\sqrt{1+y^2}-\sqrt{1-y^2}}=\frac{\sqrt{1+y^2}+\sqrt{1-y^2}}{2y^2}\tag 1$$
Then, we can develop the McLaurin Series for the numerator term on the right-hand side of $(1)$ using the Generalized Binomial Theorem, divide by $y^2$, and obtain a series expression in powers of $y$ around zero.  The leading term will be $\frac{1}{y^2}$.
And as $x\to 0$
$$\frac{1}{\sqrt{1+y^2}-\sqrt{1-y^2}}-\frac1{y^2}=-\frac{y^2}{8}+O(y^6)$$
provides an "approximation" that is of order $y^6$.
