This is an exercise in the book by Titchmarsh, 'the theory of functions', page 242.
The answer is $\frac{1}{2}\sqrt{\pi /(1-x )} $.
How to prove it?
It is a little bit surprising to me. The function $1/\sqrt{1-x} $ has a quite different Taylor expansion around $x= 0 $.
This is a Theta function. See https://en.wikipedia.org/wiki/Theta_function.
A search for "theta function asymptotics" comes up with this, titled "Asymptotics of the q-theta function", among the first hits:
www.ucs.louisiana.edu/~xxw6637/papers/CMA2009.pdf
This is the main result:
For $0 < q < 1$ and $x \in \mathbb{C}$, define $\Theta_q(x) =\sum_{n=-\infty}^{\infty} q^{k^2}x^k $, so $f(x) =(1+\Theta_x(1))/2 $.
Then, as $q \to 1^-$, $\Theta_q(x) \sim \sqrt{\dfrac{\pi}{-\ln(q)}}\exp\left(\dfrac{(\ln x)^2}{-4\ln(q)}\right) $.
Setting $x = 1$, this becomes $\Theta_q(1) \sim \sqrt{\dfrac{\pi}{-\ln(q)}} $.
If $q = 1-z$, $\Theta_{1-z}(1) \sim \sqrt{\dfrac{\pi}{-\ln(1-z)}} \sim \sqrt{\dfrac{\pi}{z}} = \sqrt{\dfrac{\pi}{1-q}} $.
Therefore
$\begin{array}\\ f(x) &=(1+\Theta_x(1))/2\\ &\sim(1+\sqrt{\dfrac{\pi}{1-x}})/2\\ &\sim\frac12\sqrt{\dfrac{\pi}{1-x}} \qquad\text{as } x \to 1\\ \end{array} $
We will use the $\theta$ function, defined by $\theta(u)=\sum_{n\in \mathbb{Z}}e^{-\pi n^2u}$ for $u>0.$ It is known that (it can be proved by a direct application of Poisson's summation formula)
$$\theta(u)=\frac{1}{\sqrt{u}}\theta\left(\frac{1}{u}\right),\ u>0\ \ \ (1).$$ If $A(u)=\sum_{n\geq 0}e^{-\pi n^2u},\ u>0,$ then it is very easy to observe that $\theta(u)=2A(u)-1.$ Consequently, $(1)$ gives
$$2\sqrt{1-e^{-\pi u}}A(u)=\sqrt{\frac{1-e^{-\pi u}}{u}}\left(2A\left(\frac{1}{u}\right)-1\right)+\sqrt{1-e^{-\pi u}}.\ \ \ (2)$$
Now, by the definition of the derivative of $u\mapsto e^{-\pi u}$ at $0,$ we have that
$$\lim_{u\rightarrow 0}\sqrt{\frac{1-e^{-\pi u}}{u}}=\sqrt{\pi}.\ \ \ (3)$$
Moreover, we have that
$$1\leq A\left(\frac{1}{u}\right) =1+\sum_{n\geq 1}e^{-\pi n^2/u}\leq 1+\sum_{n\geq 1}\int_{n-1}^ne^{-\pi x^2/u}dx\\=1+\int_0^{\infty}e^{-\pi x^2/u}dx=1+\frac{\sqrt{u}}{2},$$
where the last integral was computed by the change of variables $w=x\sqrt{\pi/u}$ and half of the known Gaussian integral. These inequalities and the squeeze theorem imply that
$$\lim_{u\rightarrow 0^+}A\left(\frac{1}{u}\right)=1.\ \ \ (4)$$
Combining $(2),(3)$ and $(4)$, we obtain
$$\lim_{u\rightarrow 0^+}\sqrt{1-e^{-\pi u}}A(u)=\frac{\sqrt{\pi}}{2}.$$
Changing variables with $x=e^{-\pi u}$ in this limit yields the desired asymptotic formula.
