Answer
$$2$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^{ - 2x}} - 1 + 2x}}{{{x^2}}} \cr
& {\text{evaluating the limit}} \cr
& = \frac{{{e^0} - 1 + 2\left( 0 \right)}}{{{{\left( 0 \right)}^2}}} \cr
& = \frac{0}{0} \cr
& \cr
& {\text{applying l'Hopital's rule twice}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{e^{ - 2x}} - 1 + 2x} \right]}}{{\frac{d}{{dx}}\left[ {{x^2}} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{e^{ - 2x}} + 2}}{{2x}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ { - 2{e^{ - 2x}} + 2} \right]}}{{\frac{d}{{dx}}\left[ {2x} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{4{e^{ - 2x}}}}{2} \cr
& = \mathop {\lim }\limits_{x \to 0} 2{e^{ - 2x}} \cr
& {\text{evaluating the limit}} \cr
& = 2{e^{ - 2\left( 0 \right)}} \cr
& = 2 \cr} $$
