Compute limit using Taylor's expansion Using Taylor’s expansion, prove that the following limit exists and compute it.

$$\lim_{x \to 0}\left(\frac {x^2}{\frac {1}{1-x} - e^x}\right)$$

In this if I am using the taylor series expansion of $e^x$ then denominator has some value as $$ 1 + (1 + x + x^2 + x^3 + \ldots) + x (1 + x + x^2 + x^3 + \ldots) $$
I am not getting how to proceed further with this.

Hint: First we have 
$$\frac{1}{1 - x} - e^x = \sum_{n=0}^\infty x^n - \sum_{n=0}^\infty \frac{x^n}{n!} =\sum_{n=0}^\infty x^n \Big(1 - \frac{1}{n!}\Big) = \frac{x^2}{2} + \frac{5x^3}{6} + O (x^4) $$
Then 
$$\lim_{x \to 0}\frac{x^2}{\frac{x^2}{2} + \frac{5x^3}{6} + O (x^4)}$$
Can you take it from here?
