By Taylor's expansion, we have

$$\begin{align*} \sqrt{1+x^2}&=1+\frac{x^2}{2}-\frac{x^4}{8} + R\left(x^4\right)\\ \cos{x}&=1 -\frac{x^2}{2!}  + R(x^2)\\e^{x^2} &= 1 + x^2  + R(x^2)\\ \sin(x^2)&=x^2  + R(x^2) \end{align*}$$

Therefore

$$\begin{align*} \frac{x^2}{2} +1 -\sqrt{1+x^2} &= \frac{1}{8}x^4 + R(x^4)\\ (\cos{x}-e^{x^2})\sin(x^2)&=\left(-\frac{3}{2}x^2 + R\left(x^2\right)\right)\left(x^2 +R\left(x^2\right)\right)=-\frac{3}{2}x^3 + R\left(x^4\right)\end{align*}$$

It follows that

$$\lim_{x\to 0}\frac{\frac{x^2}{2}+1-\sqrt{1+x^2}}{(\cos{x}-e^{x^2})\sin(x^2)}=\lim_{x\to 0}\frac{\frac{1}{8}x^4+R\left(x^4\right)}{-\frac{3}{2}x^4+R\left(x^4\right)}=\boxed{-\frac{1}{12}}$$