Answer
$-1$
Work Step by Step
$cosx=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-....$
and $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$
Plug into the limit to get
$\lim\limits_{x \to 0}\frac{1-cosx}{1+x-e^{x}}=\lim\limits_{x \to 0}\frac{1-1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-....}{1+x-1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...}$
$=\frac{1/2}{-1/2}$
$=-1$