Answer
$$ - 112$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^7} - 5{x^6} + 5{x^5} + 32}}{{x - 2}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}2{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 2} \left( {{x^7} - 5{x^6} + 5{x^5} + 32} \right) = {\left( 2 \right)^7} - 5{\left( 2 \right)^6} + 5{\left( 2 \right)^5} + 32 = 0 \cr
& \mathop {\lim }\limits_{x \to 2} \left( {x - 2} \right) = 2 - 2 = 0 \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. then differentiating the numerator and}} \cr
& {\text{denominator}} \cr
& = \mathop {\lim }\limits_{x \to 2} \frac{{{D_x}\left( {{x^7} - 5{x^6} + 5{x^5} + 32} \right)}}{{{D_x}\left( {x - 2} \right)}} \cr
& {\text{use power rule for differentiation}} \cr
& = \mathop {\lim }\limits_{x \to 2} \frac{{7{x^6} - 30{x^5} + 25{x^4}}}{1} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = 7{\left( 2 \right)^6} - 30{\left( 2 \right)^5} + 25{\left( 2 \right)^4} \cr
& = - 112 \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^7} - 5{x^6} + 5{x^5} + 32}}{{x - 2}} = - 112 \cr} $$
