Answer
$${e^4}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 2} \frac{{\int_2^x {{e^{{t^2}}}dt} }}{{x - 2}} \cr
& {\text{Using properties of limits}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{\int_2^x {{e^{{t^2}}}dt} }}{{x - 2}} = \frac{{\mathop {\lim }\limits_{x \to 2} \int_2^x {{e^{{t^2}}}dt} }}{{\mathop {\lim }\limits_{x \to 2} \left( {x - 2} \right)}} \cr
& {\text{Evaluate the limits}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{\int_2^x {{e^{{t^2}}}dt} }}{{x - 2}} = \frac{{\int_2^2 {{e^{{t^2}}}dt} }}{{\left( {2 - 2} \right)}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{\int_2^x {{e^{{t^2}}}dt} }}{{x - 2}} = \frac{{\mathop {\lim }\limits_{x \to 2} \frac{d}{{dx}}\left[ {\int_2^x {{e^{{t^2}}}dt} } \right]}}{{\mathop {\lim }\limits_{x \to 2} \frac{d}{{dx}}\left[ {x - 2} \right]}} \cr
& {\text{Differentiate, recall that }}\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)} dt} \right] = f\left( x \right) \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{\int_2^x {{e^{{t^2}}}dt} }}{{x - 2}} = \frac{{\mathop {\lim }\limits_{x \to 2} \left( {{e^{{x^2}}}} \right)}}{{\mathop {\lim }\limits_{x \to 2} \left( 1 \right)}} \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{\int_2^x {{e^{{t^2}}}dt} }}{{x - 2}} = \frac{{{e^{{{\left( 2 \right)}^2}}}}}{1} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{\int_2^x {{e^{{t^2}}}dt} }}{{x - 2}} = {e^4} \cr} $$
