Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - 1}}{x} \cr
& {\text{the limit can be written as }} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - {e^{{{\left( 0 \right)}^2}}}}}{{x - 0}} \cr
& {\text{from the definition of the derivative }}f'\left( c \right) = \mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right) - f\left( c \right)}}{{x - c}},{\text{ we compare}} \cr
& {\text{that }}\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - {e^{{{\left( 0 \right)}^2}}}}}{{x - 0}} \Rightarrow \,\,f\left( x \right) = {e^{{x^2}}}{\text{ and }}c = 0,{\text{ Thus}}{\text{,}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{{x^2}}}} \right] \cr
& f'\left( x \right) = {e^{{x^2}}}\left( {2x} \right) \cr
& and \cr
& f'\left( x \right) = 2x{e^{{x^2}}} \cr
& f'\left( 0 \right) = 2\left( 0 \right){e^{{{\left( 0 \right)}^2}}} \cr
& f'\left( 0 \right) = 0 \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - 1}}{x}{\text{ }} = 0 \cr} $$
