#### Answer

$$25\ln 5$$

#### Work Step by Step

$$\eqalign{
& \mathop {\lim }\limits_{x \to 2} \frac{{{5^x} - 25}}{{x - 2}} \cr
& {\text{rewrite}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{5^x} - 25}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{{5^x} - {{\left( 5 \right)}^2}}}{{x - 2}} \cr
& {\text{using the definition of the derivative }}f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} \cr
& {\text{comparing }}\mathop {\lim }\limits_{x \to 2} \frac{{{5^x} - {{\left( 5 \right)}^2}}}{{x - 2}}{\text{ with the definition of the derivative}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to 2} \frac{{{5^x} - {{\left( 5 \right)}^2}}}{{x - 2}}:f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to 2} \frac{{{5^x} - {{\left( 5 \right)}^2}}}{{x - 2}}:f'\left( 2 \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( 2 \right)}}{{x - 2}} \cr
& {\text{we can note that }}f\left( x \right) = {5^x}{\text{ and }}a = 2 \cr
& {\text{then}} \cr
& \,\,\,\,\,\,\,\,f'\left( x \right) = \frac{d}{{dx}}\left[ {{5^x}} \right] \cr
& \,\,\,\,\,\,\,\,f'\left( x \right) = {5^x}\ln 5 \cr
& \,\,\,\,\,\,\,\,f'\left( 2 \right) = {5^2}\ln 5 \cr
& {\text{and}} \cr
& \,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to 2} \frac{{{5^x} - 25}}{{x - 2}} = f'\left( 2 \right) = {5^2}\ln 5 \cr
& \,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to 2} \frac{{{5^x} - 25}}{{x - 2}} = 25\ln 5 \cr} $$