$\underset{x\to 0}{\lim }\frac{1-\cos 2x}{e^{x^2}-e^x+x}=$

$L=\underset{x\to 0}{\lim }\frac{e^{x^2}-e^x+x}{1-\cos 2x}$

$=\underset{x\to 0}{\lim }\frac{e^{x^2}-1+x-e^x+1}{2\left(\frac{{\sin }^{2}x}{x^2}\right).x^2}$

$=\frac{1}{2}\left[\underset{x\to 0}{\lim }\frac{e^{x^2-1}}{x^2}+\underset{x\to 0}{\lim }\frac{x-e^x+1}{x^2}\right]$

$=\frac{1}{2}\left[1+\underset{x\to 0}{Lt}\frac{1-e^x}{2x}\right]$

$=\frac{1}{2}\left[1-\frac{1}{2}\right]=\frac{1}{4}$

$\therefore \underset{x\to 0}{\lim }\frac{1-\cos 2x}{e^{x^2}-e^x+x}=4$