Consider the function
$$f(x)=\prod\limits_{n=1}^{\infty}(1x^n)$$ I am interested in an asymptotic formula for $f$ as $x\to 1$. Of course $f\to 0$ but I am interested in how fast.
Consider the function $$f(x)=\prod\limits_{n=1}^{\infty}(1x^n)$$ I am interested in an asymptotic formula for $f$ as $x\to 1$. Of course $f\to 0$ but I am interested in how fast. 


From equation (II) in Newman's A simple proof of the partition formula: $$f(z) = \sqrt{\frac{2 \pi}{1z}} \exp\left(\frac{\pi^2}{6(1z)}+\frac{\pi^2}{12} \right) (1+O(1z))$$ Note that Newman's $f$ is the reciprocal of yours; I rewrote his formula to use your $f$. This bound is inside the egg shaped region $2 z + 1z < 2$, and in particular for $z \in [0,1)$. 

